anders thorup notes on automorphic functionsweb.math.ku.dk/noter/filer/mdlar.pdf · automorphic...

Anders ThorupNotes on Automorphic Functions

Københavns Universitets Matematiske Institut1995

[Front] 2 Automorphic functions20. februar 1995

Cover figures designed in Mathematica by Jakob Dylander and Anders Jensen.

Typeset by AMS-TEX with MathTime 1.1

Automorphic functions26. februar 1995

[Toc] 1

Automorphic Functions

Copenhagen 1994

Möbius transformations (Möb)1. Möbius transformations2. Fixed points3. Non-Euclidean plane geometry4. Proper actions

Discrete subgroups (Discr)1. Isotropy groups and their generators2. Properly discontinuous actions3. Finite normal fundamental domains4. Canonical fundamental domains5. The Euler characteristic

Modular groups (Mdlar)1. Finite projective lines2. Small modular groups3. Congruence subgroups4. Some modular fundamental domains

Automorphic forms (Autm)1. Automorphic factors2. Examples I3. The signs of a factor4. Automorphic forms5. Fourier expansions6. The Main Theorems7. Examples II8. The Proofs

Poincaré Series and Eisenstein Series (Poinc)1. Poincaré series2. A Fourier expansion of an Eisenstein Series3. Example: Eisenstein series for the theta group

Appendix (App)1. Bernoulli and Euler numbers2. Fourier expansion of Eisenstein series3. The function of Weierstrass4. Notes on elliptic curves

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[Toc] 2 Automorphic functions26. februar 1995

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[Möb] 1

Möbius transformations

1. Möbius transformations.

(1.1) Setup. The general linear group GL2(C) consists of all 2 × 2 matrices with complexentries and non-zero determinant,

α =[a b

c d

]where ad − bc �= 0. (1.1.1)

The special linear group SL2(C) is the subgroup formed by matrices (1.1.1) for which thedeterminant, det α = ad − bc, is equal to 1. For a matrix α of the form (1.1.1) and a point zof the Riemann Sphere (the extended complex plane C = C ∪ {∞}), we define the product,

α · z := az + b

cz + d. (1.1.2)

The transformation z �→ α · z of the Riemann sphere is called the Möbius transformationassociated to the matrix α. The denominator of the fraction (1.1.2) will play an importantrole; we define, as a function of the matrix α and the complex number z,

J (α, z) := cz + d. (1.1.3)

Denote by C2 the 2-dimensional vector space of columns, and by (C2)∗ the subset ofnon-zero columns. Then there is a surjective map (C2)∗ → C defined by

[z1

z2

]�→ z1/z2.

The non-zero columns that are mapped to a given point z of C are called the representatives ofz. Clearly, the representatives of a given point form the non-zero columns in a 1-dimensionalvector subspace of C2.

Three matrices deserve a special notation:

s :=[ 0 −1

1 0

], t :=

[ 1 10 1

], u :=

[ −1 −11 0

].

Note that s = tu.

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[Möb] 2 Automorphic functions26. februar 1995

(1.2) Example. The translation z �→ z + b and the multiplication z �→ az for a �= 0 areMöbius transformations, associated to the matrices,

[ 1 b

0 1

],

[a 00 1

].

The point ∞ is fixed. On the other hand, if d is given complex number, then d is mapped to∞ under the Möbius transformation associated to the matrix,

[ 0 1−1 d

].

(1.3) Proposition. (1) Two matrices α and β define the same Möbius transformation, if andonly if they are proportional.

(2) The following equations hold for the product defined in (1.1.2):

α · (β · z) = (αβ) · z, 1 · z = z.

(3) Given in C two sets of 3 different points, (u, v, w) and (u′, v′, w′). Then there is aMöbius transformation z �→ α · z under which (u, v, w) �→ (u′, v′, w′), and the matrix α isunique up to multiplication by a non-zero scalar. In particular, the Möbius transformationz �→ α · z is unique.

(4) If a column z represents the point z, then the column αz represents the image pointα · z.Proof. Clearly (4) holds, and (2) is an immediate consequence. In (1), obviously, if the matri-ces α and β are proportional, then they define the same Möbius transformation. Conversely,it follows from (3) that if two matrices define the same Möbius transformation, then they areproportional.

To prove (3), consider a set of 3 columns (u, v, w) representing 3 different points (u, v, w)in C. Say that the set of representatives is balanced, if

w = u+ v. (1.3.1)

When a representative w of w is given, there is a unique choice of representatives of u and v,such that the resulting set of representatives is balanced. Indeed, the vectors representing agiven point form the non-zero vectors in a 1-dimensional vector subspace of C2. Therefore,since C2 is a 2-dimensional vector space, the decomposition (1.3.1) is the unique decompo-sition of the vector w into a sum of two vectors lying in two given different 1-dimensionalsubspaces of C2. It follows in particular that a balanced set of representatives is unique up tomultiplication by a non-zero scalar.

Choose two balanced sets of representatives, (u, v, w) for (u, v, w), and (u′, v′, w′) for(u′, v′, w′). If (u, v, w) is mapped to (u′, v′, w′) under a Möbius transformation z �→ α · z,then, by (4), (αu, αv, αw) is a set of representatives for (u′, v′, w′), and it is a balanced set,

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[Möb] 3

because (u, v, w) is balanced. Therefore, by uniqueness of balanced sets, the set (αu, αv, αw)is proportional to the set (u′, v′, w′). Hence we obtain the matrix relation,

α(u, v, w) ∼ (u′, v′, w′). (1.3.2)

In particular, extracting the equations of the first two columns, we obtain that

α ∼ (u′, v′)(u, v)−1. (1.3.3)

Conversely, if α is defined by equality in (1.3.3), then equality holds in (1.3.2), since, on bothsides, the third column is the sum of the first and the second. Therefore, by (4), the Möbiustransformation associated to α maps (u, v, w) �→ (u′, v′, w′).

Thus (3) has been proved, and the proof is complete.

(1.4) Definition. Usually we writeαz for the productα·z. It follows from Proposition (1.3)(2)that the product (α, z) �→ αz defines an action of the group GL2(C) on the Riemann sphereC. Clearly, the Möbius transformations are analytic automorphisms. Hence the associatedrepresentation is a homomorphism of groups,

GL2(C) → Autan(C).

It is well known that the homomorphism is surjective, that is, every analytic automorphismof the Riemann sphere is a Möbius transformation. It follows from Proposition (1.3)(1) thatthe kernel of the homomorphism is the subgroup C∗ of non-zero scalar matrices. For anysubgroup G of GL2(C), we denote by PG the quotient of G modulo the subgroup of scalarmatrices contained in G, or equivalently, PG is the image of G in the group of Möbiustransformations. The subgroupG is called inhomogeneous if it contains no non-trivial scalarmatrix, that is, if G = PG.

Note that PGL2(C) is the group of all Möbius transformations. The group PSL2(C) is thequotient,

PSL2(C) = SL2(C)/± 1.

Every matrix in GL2(C) is proportional to a matrix in SL2(C), as it follows by dividing bya square root of the determinant. Hence PSL2(C) = PGL2(C), and every Möbius transfor-mation is associated to a matrix in SL2(C). A subgroupG of SL2(C) is homogeneous if andonly if it contains the matrix −1.

(1.5) Example. To determine the Möbius transformation under which (∞, 0, i) is mappedto (1,−1, 0), consider these two sets of balanced representatives for the two sets of points:[

i 0 i

0 1 1

],

[ 1 −1 01 1 2

].

It follows from (1.3)(3) that the Möbius transformation is associated to the matrix,[ 1 −11 1

][i 00 1

]−1 ∼[ 1 −i

1 i

].

Hence the transformation is the map z �→ (z−i)/(z+i). It is called the Cayley transformation.To obtain a matrix in SL2(C), divide the matrix by the square root of its determinant 2i, thatis, divide the matrix by 1 + i.

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(1.6) Corollary. Any Möbius transformation preserves the cross ratio,

df(u, v, w, z) := u− w

v − w

/u− z

v − z. (1.6.1)

In particular, a Möbius transformation preserves angles, it maps a circle to a circle and theinterior of an oriented circle to the interior of the image circle.

Proof. The cross ratio of four different points in C is the complex number defined by (1.6.1).It is easy to extend the definition to the case when one of the four points is equal to ∞. Clearly,for any set of four columns (u, v, w, z) representing (u, v, w, z), we have the equation,

df(u, v, w, z) = det(u, w)

det(v, w)

/det(u, z)

det(v, z). (1.6.2)

The first assertion of the Corollary is the equation df(αu, αv, αw, αz) = df(u, v, w, z) forany Möbius transformation z �→ αz. By (1.3)(4), the first assertion follows from (1.6.2),since det(αu, αv) = det α det(u, v).

To prove the remaining assertions, fix a point v in C and, in C, a point u �= v. For anypoint w in C different from u and v, consider the circle Cw through u, v, and w, oriented bythe order uvw. Let tw be the oriented tangent in the point v to the circle Cw. If z is a fourthpoint, and tz is defined similarly from the circle Cz, then the following equation holds for theangle from tw to tz:

� (tw, tz) = arg df(u, v, w, z). (1.6.3)

To prove Equation (1.6.3), consider first the case when u = ∞. Then tw and tz are the twooriented straight lines from v to w and z, and the left hand side is the angle between them.On the other side, the cross ratio reduces to the ratio (v − z)/(v − w). Hence the argumentof the cross ratio is equal to the left hand side. Assume next that u �= ∞, and consider theoriented (straight) line t = −→uv from u to v. It is the common chord to the two circles Cwand Cz. It follows from elementary plane geometry properties of the circle Cw, that the anglefrom −→wv to −→wu is equal to the angle from tw to t ,

� (tw, t) = � (−→wv,−→wu).

Moreover, the angle on the right hand side is equal to the argument of (u−w)/(v−w). Now,the equation (1.6.3) follows from the additivity � (tw, t)+ � (t, tz) = � (tw, tz).

As a consequence of (1.6.3), the cross ratio df(u, v, w, z) belongs to R+ if and only if z ison the arc vu of the circle Cw, it belongs to R− if and only if z is on the arc uv of the circle,it belongs to the upper half plane if and only if z is an inner point of the circle Cw, and itbelongs to the lower half plane if and only if z is exterior to the circle.

Therefore, as a Möbius transformation preserves the cross ratio, it follows that circles andinteriors of circles are preserved, and it follows from (1.6.3) that angles between circles arepreserved.

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[Möb] 5

(1.7) Remark. In the proof of (1.6) we obtained information on the argument of the crossratio df(u, v, w, z) by writing the cross ratio as a quotient of two fractions,

u− w

v − w,

u− z

v − z. (1.7.1)

Similarly, we can obtain information on the modulus of the cross ratio. Assume for simplicitythat the three points u, v, and w are not on a straight line (in particular, they are differentfrom ∞), and consider the (ordinary) triangle defined by them. Denote by W the angle at wand by Uw and Vw the angles at u and v. The two lengths |u − w| and |v − w| are sides ofthe triangle, and hence the quotient |u− w|/|v − w| is equal to the quotient sinVw/ sinUw.Hence, as the sum of the three angles is equal to π , we obtain for the modulus of the firstfraction (1.7.1) the equation,

∣∣∣u− w

v − w

∣∣∣ = sinVwsinUw

= cosW + cotUw sinW. (1.7.2)

Thus the modulus of the cross ratio df(u, v, w, z) is the quotient of the expression (1.7.2) andthe expression obtained similarly replacing w by z.

Fix u and v and an oriented circle C through u and v. Consider the expression (1.7.2) asa function dC(w) defined for points w different from u and v on the circle C. The angle Wis constant, say equal to θ , on the arc from v to u, and on the arc from u to v the angle Wis equal to π − θ . Consider points on the arc from v to u. Then, as w runs from v to u, theangle Uw increases from 0 to π − θ , and consequently, the function dC(w) decreases from+∞ to 0.

Consider, for 4 different points (u, v, w, z) on C, the cross ratio df(u, v, w, z). It followsfrom the proof of Corollary (1.6), that the cross ratio is real and positive if and only ifw and zbelong to the same of the two arcs determined by u and v. In particular, whenw and z belongto the same arc, then the cross ratio is equal to its modulus, and hence equal to dC(w)/dC(z).As a consequence, when z is on the arc from v to u containingw, the cross ratio df(u, v, w, z)increases from 0 to 1 as z runs from v to w, and it increases from 1 to +∞ as z runs from w

to u.It is easy to prove that the latter assertion holds also when the circle is a straight line.

(1.8) Definition. An open disk in C will simply be called a disk. Thus a disk is either anopen half plane in C, or the interior of a usual circle in C, or the exterior (including ∞) ofa usual circle. The boundary ∂D of a disk D is a circle, always oriented counter clockwisearound the disk. If D is a disk, we denote by SL(D) the stabilizer of D in SL2(C), that is,the subgroup of SL2(C) consisting of matrices α for which αD = D.

Throughout, we denote by H the upper half plane: z > 0, and by E the open unit disk:|z| < 1. The boundary (in C) of H is the extended real line R, and the boundary of E is theunit circle: |z| = 1.

(1.9) Corollary. Given two triples (D, w, u) and (D′, w′, u′), each consisting of a disk, apoint in the disk, and a point on the boundary of the disk. Then there is a unique Möbiustransformation mapping the first triple to the second.

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Proof. The point w is in D and the point u is on the boundary ∂D. Clearly, there is a uniquecircle C orthogonal to ∂D, and passing through w and u. The circles C and ∂D intersectin two points, one of which is u. Denote by v the second point of intersection. Define C ′and v′ similarly from the second triple. Clearly, for a Möbius transformation as required, theboundary ofD is mapped to the boundary ofD′. Hence the circleC is mapped to the circleC ′,and consequently, v is mapped to v ′. Thus, by Proposition (1.3), the Möbius transformationis the unique transformation under which

(u, v, w) �→ (u′, v′, w′). (1)

Conversely, under the transformation determined by (1), the circle C is mapped to the circleC ′. Hence the boundary of D, which is orthogonal to C, is mapped to the boundary of D ′,and hence D is mapped to D′. Thus the Möbius transformation determined by (1) has therequired properties.

(1.10) Example. Clearly, the Cayley transformation of (1.5) is the unique Möbius transfor-mation mapping (H, i,∞) onto (E, 0, 1).

(1.11) Remark. Let α be a matrix of GL2(C). The transformation z �→ αz, where z denotesthe complex conjugate of z, is called the anti-transformation associated to α. Note thatan anti-transformations is not a Möbius transformation. However, the composition of anti-transformations, associated to matrices α and β, is a Möbius transformation, associated tothe product αβ.

Clearly, under an anti-transformation, the cross ratio of four points is changed into thecomplex conjugate. As a consequence, an anti-transformation preserves circles, but anglesbetween circles are reversed. The interior of an oriented circle is mapped to the exterior ofthe image circle (when the image circle is given the image orientation).

(1.12) Example. Complex conjugation is the anti-transformation associated to the identitymatrix. Under complex conjugation, the unit disk E is mapped to itself, but the orientationof the boundary is reversed.

(1.13) Exercise. Prove for z ∈ H and σ ∈ SL2(R) that (σz) = |J (σ, z)|−2z.

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[Möb] 7

2. Fixed points.

(2.1) Definition. Let α be a matrix in GL2(C), and consider the associated Möbius transfor-mation z �→ αz. Clearly, a point z of C is a fixed point of the transformation, if and onlyif the representatives of z in (C2)∗ are eigenvectors of α. Therefore we get the followingclassification of transformations:

Case 1. The matrixα has one eigenvalue and the corresponding eigenspace is of dimension2. In this case, the matrix α is a scalar matrix,

[λ 00 λ

].

The associated transformation is the identity, and every point in C is fixed.Case 2. The matrix α has exactly one eigenvalue λ, and the eigenspace is of dimension

1. In this case, the transformation has exactly one fixed point. The transformation (and thematrix) is called parabolic. The matrix is similar to a matrix of the form,

[λ b

0 λ

],

where b �= 0.Case 3. The matrix α has two different eigenvalues λ1 and λ2 (necessarily both with a one

dimensional eigenspace). In this case, the transformation has two fixed points. The matrix αis similar to a matrix for the form, [

λ1 00 λ2

].

Of particular geometric interest is the quotient λ1/λ2 of the two eigenvalues (changed to itsinverse when the two eigenvalues are interchanged). The matrix α is called hyperbolic, ifthe quotient is real and positive, and elliptic, if the quotient is of modulus 1. It is calledloxodromic if it is neither elliptic nor hyperbolic.

Note that the quotient of the two eigenvalues of a matrix α is equal to 1, if and only thematrix is either a scalar or a parabolic matrix.

(2.2) Example. Obviously, under the action of SL2(C) on C, the isotropy group of the point∞ is the subgroup formed by the following matrices:

[a b

0 a−1

]. (2.2.1)

Clearly, among the matrices in (2.2.1) the parabolic matrices are the following:

±[ 1 b

0 1

];

their associated transformations are translations z �→ z + b for b �= 0.

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Consider the matrices in (2.2.1) that fix in addition the point 0. Of these matrices, thehyperbolic matrices are the following:

±[r 00 r−1

],

where r ∈ R+ − {1}; their associated transformations are multiplications z �→ r 2z. Theelliptic matrices are the following: [

eiθ 00 e−iθ

],

where θ ∈ R − Zπ ; their associated transformations are rotations z �→ e2iθ z.An invariant closely related to the geometric properties of Möbius transformations is the

following, defined for any matrix α ∈ GL2(C):

h(α) := 1

2

tr(α2)

det α.

By Proposition (1.3)(1), the invariant h(α) depends on the associated Möbius transformationz �→ αz only. Moreover, the invariant is unchanged if α is replaced by a conjugate σασ−1.Expressed by the eigenvalues in (2.1), we have that h(α) = 1

2 (λ1/λ2 + λ2/λ1). Thus thequotient λ1/λ2 and its inverse λ2/λ1 are the two roots of the quadratic polynomial,

λ2 − 2 h(α)λ+ 1.

In particular, if α is not a scalar matrix, then α is parabolic, if and only if h(α) = 1, α ishyperbolic if and only if h(α) is real and in the interval 1 < h < +∞, and α is elliptic, ifand only if h(α) is real and in the interval −1 ≤ h < 1.

(2.3) Lemma. (1) The stabilizer SL(H) of the upper half plane H is the subgroup SL2(R)consisting of matrices,

[a b

c d

]where a, b, c, d ∈ R and ad − bc = 1.

The isotropy group SL(H)∞ of the point ∞ on the boundary of H is the subgroup consistingof matrices, [

a b

0 a−1

]for a ∈ R∗, b ∈ R. (2.3.1)

In particular, the isotropy group SL(H)∞ is non-compact and non-commutative. The follow-ing relation holds, [

a b

0 a−1

][ 1 h

0 1

][a b

0 a−1

]−1 =[ 1 a2h

0 1

]. (2.3.2)

Moreover, the subset of SL(H)∞ consisting of matrices that are either parabolic or ±1 is asubgroup, isomorphic to {±1} × R.

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[Möb] 9

(2) The stabilizer SL(E) of the unit diskE is the subgroup SU1,1(C) consisting of matrices,[a b

b a

]for a, b ∈ C and |a|2 − |b|2 = 1.

The isotropy group SL(E)0 of the point 0 in E is the subgroup consisting of matrices,[a 00 a

]for a ∈ C, |a| = 1. (2.3.3)

In particular, the isotropy group is compact and commutative (and isomorphic to the unitcircle U1(C)).

Proof. (1) Let α be a matrix in SL(H). The Möbius transformation z �→ αz maps the diskH onto itself and, consequently, it maps the boundary R onto itself. Therefore, by Section1, the Möbius transformation is associated to a matrix α ′ with real entries. Since α′ has realentries, the imaginary part of α′ · i is equal to a positive scalar times the determinant of α ′.It follows that the determinant is positive, and dividing the matrix by a square root of thedeterminant, we may assume that α ′ has determinant 1. As α and α′ define the same Möbiustransformation, it follows that α = ±α ′. Hence α belongs to SL2(R). Conversely, it isobvious that any matrix in SL2(R) belongs to SL(H).

The remaining assertions of (1) and the similar assertions of (2) are left as an exercise.

(2.4) Exercise. (1) Prove that the subgroup SL(H)i of matrices in SL(H) having i as fixedpoint is the subgroup SO2(R) of matrices,[

a b

−b a

]for a, b ∈ R and a2 + b2 = 1.

Prove that the subgroup SL(H)1,−1 of matrices in SL(H) having 1 and −1 as fixed points isthe subgroup SO1,1(R) of matrices,[

a b

b a

]for a, b ∈ R and a2 − b2 = 1.

(2.5) Corollary. Consider a disk D in C. No matrix in SL(D) is loxodromic. Let σ �= ±1be a matrix in SL(D). If σ is parabolic, then the fixed point of σ belongs to the boundary∂D. If σ is elliptic, then σ has one fixed point in D and the other fixed point belongs to thecomplement of the closure of D. Finally, if σ is hyperbolic, then the two fixed points of σbelongs to the boundary of D.

Proof. After conjugation, we may assume that the disk is the upper half plane H. Then, byProposition (2.3), the matrix σ has real entries. Hence, the eigenvalues of σ are either realor a pair of complex conjugate numbers. Clearly, in the first case there are real eigenvectorsand, consequently, either σ is parabolic with a fixed point in R or σ is hyperbolic withtwo fixed points in R. In the second case, if a column is an eigenvector corresponding toone eigenvalue, then the conjugate column is an eigenvector corresponding to the complexconjugate eigenvalue. Hence, in the second case, σ has one fixed point in the upper half planeH and the complex conjugate fixed point in the lower half plane.

Thus the assertions hold.

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(2.6) Note. By Corollary (1.9), for any two disks D and D′ there is a Möbius transformationz �→ αz defining an isomorphism,

α : D′ → D,

of D′ onto D. In addition, for a given pair of points u ∈ D and u′ ∈ D′ (or u ∈ ∂D andu′ ∈ ∂D′), we may choose α such that αu′ = u.

Under the isomorphism α, points z ∈ D correspond to points zα := α−1z in D′, functionsf defined on D correspond to functions f α := f α on D′, and automorphisms σ of Dcorrespond to automorphisms σ α := α−1σα on D′. The latter correspondence extendsto matrices: matrices σ in SL(D) correspond to matrices σ α := α−1σα in SL(D′). Thecorrespondence is called conjugation. In particular,

SL(D′) = SL(D)α,

is the conjugate subgroup of SL(D). Clearly, the property of being a fixed point is preservedunder conjugation, and for the isotropy groups we obtain the equation,

SL(D′)u′ = SL(D) αu ,

when u′ = uα .Usually, of the two disks in (1.8), we take (E, 0) as the model of a disk D and a point

u ∈ D, and we take (H,∞) as the model of a disk D and a point u ∈ ∂D.A property of points u ∈ D ∪ ∂D will often be defined by first defining the property for

0 ∈ E and ∞ ∈ ∂H and then in general by choosing an isomorphism σ from (D, u) to one ofthe standard models. In these cases, the property will be said to be defined by conjugation.In each case, the definition has to be justified by proving that the property is independent ofthe choice of conjugation σ .

(2.7) Definition. Let D be a disk, and let G be a subgroup of SL(D). By definition, aG-elliptic point is a point u in D which is fixed under some nontrivial (necessarily elliptic)matrix in G, and a G-parabolic point is a point u of the boundary ∂D which is fixed undersome parabolic matrix inG. A point of D which is notG-elliptic may be called aG-ordinarypoint.

As a subgroup of SL(D), the group G acts on D and on the boundary ∂D. Clearly, apoint u of D is G-elliptic if and only if the isotropy group Gu is non-trivial (that is, containsa matrix different from ±1), and a point u of ∂D is G-parabolic, if and only if the isotropygroup Gu contains a parabolic matrix.

Note that the properties are preserved under conjugation. If D′ = αD, thenG′ := αGα−1

is a subgroup of SL(D′), and for points u and u′ := αu, where u is in D or ∂D, we have thatG′u′ = αGuα

−1. Moreover, the point u is G-parabolic or G-elliptic respectively, if and onlyif u′ is G′-parabolic or G′-elliptic.

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[Möb] 11

3. Non-Euclidean plane geometry.

(3.1) Lemma. Let C be an ordinary circle in C, with its center on the positive real axisand such that C is orthogonal to the unit circle around 0. Let r < r ′ be the two points ofintersection of C and the real axis. Let l be a straight line through 0 intersecting the circleC in two points z and z′ with z ≤ z′. Denote by A the angle between l and the real axisand denote by B the angle between l and the tangent in z to the circle C. Then,

cosA = |z|−1 + |z|r−1 + r

, sinB = |z|−1 − |z|r−1 − r

. (3.1.1)

Proof. Let z0 denote the midpoint of the chord zz′. Then z0 = (z′ + z)/2. In particular,r0 = (r ′ + r)/2 is the center of the circle C. Clearly, the straight line from r0 to z0 isorthogonal to the line l. Moreover, the angle at r0 between the lines r0z and r0z0 is equal toB. Hence we have the equations,

cosA = |z0|r0

, sinB = |z0 − z|r0 − r

. (3.1.2)

As is well known, the product of distances, |z||z′|, is independent of l. By hypothesis, if lis the tangent to the circle C, then z = z′ is on the unit circle, and hence |z||z′| = 1. Hence|z||z′| = 1 for an arbitrary line l intersecting the circle. In particular, rr ′ = 1. Therefore, theequations (3.1.2) imply the equations (3.1.1)

(3.2) Setup. Consider for the rest of this section a diskD. By definition, a line inD is (the partin D of) a circle orthogonal to the boundary of D. The circle will intersect the boundary of Din two points, called the limit points of the line. By Corollary (1.6), a Möbius transformationα : D → D′ maps lines of D to lines of D′. As a consequence, assertions about points, lines,and limit points in D hold in general, if they hold for one of the two standard disks, H andE. For instance, in the unit disk E, the lines through the point 0 are the diameters of the unitcircle. Hence, for every point z �= 0 in E there is a unique line of E passing through 0 andz. As a consequence, for any two different points z and w in D, there is a unique line of Dpassing through z and w. Similarly, for any point w in D and any line l not passing throughw, there is a unique point z on l such that the line through w and z is orthogonal to l.

(3.3) Definition. Consider the line in D passing through two different points w and z of D.The line has two limit points u, v on the boundary of D; we may choose the notation suchthat the open arc wzu of the line is contained in D. It follows from Remark (1.7) that thecross ratio df(u, v, w, z) is real and greater than 1. The number,

distD(w, z) := log df(u, v, w, z),

which is positive, is called the non-euclidean distance between w and z. Note that if w = z,then the cross ratio on the right hand side is equal to 1 for any pair of different points u and

15


v on the boundary of D. Accordingly, as might be expected, the distance is defined to be 0when w = z.

The non-euclidean distance is a metric in D, that is, the triangle inequality holds for any 3points w, x, z in D. This fact, and the fundamental trigonometric formulas of non-euclideanplane geometry will be proved in the following.

Note that any Möbius transformation α : D → D′ is an isometry with respect to thedistances of D and D′. In particular, the matrices in SL(D) define isometries of the disk D.

(3.4) Remark. If an oriented line l is given in D then there is a signed distance dist l (w, z)defined for points w and z on l as follows: Let u and v be the limit points of l, chosen suchthat the arc from v to u is the part of l in D. Then distl (w, z) := log df(u, v, w, z). Clearly,for any 5 different points in C we have the equation,

df(u, v, w, z) = df(u, v, w, x) df(u, v, x, z).

Hence, for 3 points w, x, z on a line l, we have the additivity for the signed distance,

distl(w, z) = distl (w, x)+ distl(x, z).

In particular, distl(w, z) = − distl (z, w).

(3.5) Example. In the unit diskE, the distance from 0 to a point z inE is given by the formula,

dist(0, z) = log1 + |z|1 − |z| . (3.5.1)

Indeed, the two sides of the formula are unchanged under a rotation around 0. Hence we mayassume that the point z is real and 0 < z < 1. Then the line through 0 and z is the real axis,and the limit points are u = 1 and v = −1. The formula is now obvious.

(3.6) Remark. It follows from the formula (3.5.1) for the unit disk E that a non-euclideancircle (also called a geodesic circle) with center 0, that is, the set of points in E of a fixed(non-euclidean) distance to 0, is an ordinary circle. Moreover, the line in E from the center 0to a point on a geodesic circle around 0 is orthogonal to the circle. Similarly, an open geodesicdisk around 0, that is, a set of points in E whose distance from 0 is strictly less than a givenpositive number, is an ordinary disk contained in E, and the system of geodesic disks around0 form a basis for the system of neighborhoods of 0. As a consequence, for any point w inan arbitrary disk D, the geodesic circles in D around w are (ordinary) circles, and the linein D from w to a point on a geodesic circle around w is orthogonal to the circle. Similarly,geodesic disks in D are ordinary disks contained in D, and the system of geodesic disksaround w is a basis for the system of neighborhoods of w. In other words, the non-euclideandistance induces a topology in D equal to the ordinary topology of D as a subspace of C.

(3.7) Example. Consider the upper half plane H, and two points z and w in H. Then thedistance from w to z is given by the formula,

cosh dist(w, z) = 1 + |w − z|22zw . (3.7.1)

16


[Möb] 13

Indeed, assume first that the two points have different real part. Then the line in H from w

to z is a circle orthogonal to the real axis. The circle intersects the real axis in two points,labeled u, v in the usual order, see (3.3). Let r be the radius of the circle, and o its center. Letθw be the angle vuw and θz the angle vuz. Then, clearly,

tan θw = |v − w||u− w| and tan θz = |v − z|

|u− z| .

As the cross ratio df(u, v, w, z) is positive real, it is equal to tan θz/ tan θw. Therefore, theleft hand side of (3.7.1), is equal to the expression,

1

2

( tan θztan θw

+ tan θwtan θz

). (3.7.2)

On the other side, the angles vow and voz are, respectively, 2θw and 2θz. Hence, the anglewoz is equal to 2(θz − θw). Therefore,

sin 2θw = wr, sin 2θz = z

r, sin(θz − θw) =

12 |z − w|

r.

From these equations it follows that the right hand side of (3.7.1) is equal to the expression,

1 + 2 sin2(θz − θw)

sin 2θw sin 2θz. (3.7.3)

By elementary trigonometric formulas, the expressions (3.7.2) and (3.7.3) are equal. Hence(3.7.1) holds.

The equation (3.7.1) is easily seen to hold when z and w have the same real part. Hencethe equation holds in general.

(3.8) Exercise. Consider the non-euclidean distance in H. Prove for two pointsw and z withthe same real part that

dist(w, z) = | log(z/w)|. (3.8.1)

Prove for h positive and real that

dist(z, z+ h) = 2 log

(h

2z +√( h

2z)2 + 1

). (3.8.2)

Note in particular that the distance converges to zero for z → ∞.

(3.9) Setup. Consider in D a triangle A,B,C, that is, A, B and C are three different pointsof D, in general assumed to be not on the same line of D. Denote by a, b, and c the sides ofthe triangle, that is, a is the line through B and C, b is the line through A and C, and c is theline through A and B. It is customary to denote by the same symbols also the lengths of thesides of the triangle, that is, a is the distance distD(B, C) etc. Similarly, A will also denotethe (unoriented) angle at A, that is, the angle between the lines b and c, etc.

17


(3.10) Proposition. Assume in the setup of (3.9) that the angle at C is a right angle. Thenthe following formulas hold:

cosA = tanh b

tanh c, sinA = sinh a

sinh c, cosh c = cosh a cosh b. (3.10.1)

Proof. To prove the first formula, we may after conjugation assume that D is the unit diskE and A is the point 0. Moreover, after a rotation around 0 we may assume that C is a realpoint r with 0 < r < 1. Then the line a is an ordinary circle with center on the positive realaxis and orthogonal to the unit circle, and B is a point z on the line a. The distances b and cfrom 0 to C and from 0 to B are, by Example (3.5), given by exp b = (1 + r)/(1 − r) andexp c = (1 + |z|)/(1 − |z|). Therefore, the first formula of (3.10.1) follows from the firstformula of (3.1.1).

Similarly, the second formula, in the symmetric form sinB = sinh b/ sinh c, follows fromthe second formula of (3.1.1). Finally, the third formula of (3.10.1) follows from the first twoby using the relation cos2 A+ sin2A = 1.

(3.11) Proposition. In the setup of (3.9), we have for an arbitrary triangle ABC in D thecosine relation:

cosh c = cosh a cosh b − sinh a sinh b cosC,

and the sine relations:sinA

sinh a= sinB

sinh b= sinC

sinh c.

Proof. Denote byH the point on the side a such that the line from A toH is orthogonal to a.Then the triangles ACH and ABH have a right angle at H and they have as common sidethe line h from A to H . Denote by x the side CH of ACH and by y the side BH of ABH .

By the third formula of (3.10.1), applied to ACH and ABH , we obtain that cosh h =cosh b/ cosh x = cosh c/ cosh y. Hence,

cosh c = cosh bcosh y

cosh x. (1)

Assume that the pointH on a lies between B and C (the two alternative cases are left to thereader). Then y = a − x by (3.4). Hence, by the addition formula for the hyperbolic cosine,we obtain from (1) that,

cosh c = cosh a cosh x − sinh a sinh x

cosh xcosh b = cosh a cosh b − sinh a cosh b tanh x. (2)

Moreover, by the first equation of (3.10.1), we have that tanh x = tanh b cosC. Hence theequation (2) implies the cosine relation.

To prove the sine relation, note that the second formula of (3.10.1) implies the equationssinC sinh b = sinh h = sinB sinh c. Hence the second sine relation holds, and by symmetrythey all hold.

18


[Möb] 15

(3.12) Remark. The triangle inequality for the non-euclidean distance follows from thecosine relation. Indeed, the right hand side of the relation is at most equal to cosh a cosh b+sinh a sinh b = cosh(a + b), and consequently c ≤ a + b. Moreover, equality holds if andonly if the angle C is equal to π , that is, if and only if C belongs to the line segment from A

to B.

(3.13) Setup. In the geometric language, the elliptic transformations ofD are called rotations,the hyperbolic transformations are called translations, and the parabolic transformations arecalled limit rotations. A translation has two limit point as fixed points; the line between themis called the axis.

We will study these maps in more detail in the following. In addition we will need thereflections: Let l be a line in D. Choose a Möbius transformation z �→ αz mapping the unitdisk E onto the given disk D and mapping the real axis of E to the given line l of D. Thenthe reflection in l is the transformation

ρl := α( )cα−1,

where ( )c denotes complex conjugation. Note that the resulting transformation of D isindependent of the choice of α. A different choice of α would be of the form αβ where β isa matrix in SL(E) leaving invariant the real axis. Thus β has real entries, and consequentlyβ commutes with complex conjugation ( )c.

(3.14) Lemma. (1) Let z �→ σz be a rotation in D with fixed point w. Then every line tthrough w is mapped to a line through w, and the angle from t to σ t is a constant, given bythe formula,

cos � (t, σ t) = h(σ ).

Moreover, if t and t ′ are oriented lines through a given pointw, then there is a unique rotationaround w that maps t to t ′. (2) Let z �→ σz be a translation in D with axis l. Then everypoint z in l is mapped to a point in l, and the distance from z to σz is a constant, given by theformula,

cosh dist(z, σz) = h(σ ).

Moreover, if z and z′ are given points on a line l, then there is a unique translation with axisl that maps z to z′.

Proof. (1) After conjugation, we may assume that D is the unit disk E and that w = 0. Thenσ is a matrix,

σ =[eiθ 00 e−iθ

],

and the associated Möbius transformation is the ordinary rotation z �→ e2iθ z. Hence thefirst and the last assertions of (1) hold, and the angle from t to σ t is equal to 2θ . Moreover,h(σ ) = 1

2 tr(σ 2) = cos(2θ), and hence the formula of (1) holds.

19


(2) After conjugation, we may assume that D is the upper half plane H and that the axis lis the imaginary axis. Then σ is a matrix,

σ =[r 00 r−1

],

and the associated Möbius transformation is the multiplication z �→ r 2z. Hence a point iy onthe line l is mapped to the point ir 2y on l, and the distance from iy to ir2y is, by (3.8.1), equalto | log(r2y/y)| = | log r2|. Hence the first and the last assertions of (2) hold. Moreover,h(σ ) = 1

2 tr(σ 2) = (r2 + r−2)/2, and hence the formula of (2) holds.

(3.15) Definition. For a point P in D, denote by δP the half turn around P , that is, thenon-euclidean rotation with angle π around P . For a line l in D, denote by ρl the reflectionin l (Note that ρl is not a Möbius transformation, it is an anti-analytic automorphism of C).Moreover, for two different points P and Q, denote by τQP the translation along the linethrough P and Q that maps P to Q. Finally, for two different oriented lines l and m thatintersect at a point P , denote by δml the rotation around P that maps l to m.

(3.16) Lemma. (1) If P and Q are different points of D, then the we have the equation,

τ 2QP = δQδP . (3.16.1)

(2) If l and m are different oriented lines that intersect at a point P in D, then we have theequation,

δ 2ml = ρmρl. (3.16.2)

(3) Ifm and l are two different lines with no point of intersection in D or in ∂D, then we havethe equation,

τ 2QP = ρmρl, (3.16.3)

where P and Q are the unique points of l and m such that the line through P and Q isorthogonal to l and m.

Proof. (1) Let l be the oriented line from P toQ and let u and v be its limit points. Clearly uand v are interchanged by any of the half turns δP and δQ. Therefore u and v are fixed pointsfor the composition δQδP . Hence the composition is a translation with axis l. Obviously thesquare τ 2

QP is a translation with axis l. Hence, to prove Equation (3.16.1), it suffices to showthat there is one point at which the two sides of the equation takes the same value. Let R bethe point on the line l for which the pointQ is the midpoint of the line segment PR. Clearly,the pointQ is mapped to R by the translation τQP , and P is mapped to R by the half turn δQ.It follows that the point P is mapped to R by any of the two sides of the equation (3.16.1).Therefore, the equation holds.

(2) The composition ρmρl of anti-transformations is a Möbius transformation, and it hasobviously the pointP as fixed point. Hence the composition is a rotation aroundP . Obviously,under the composition, the line l is mapped to the line l ′ = ρml through P for which the

20


[Möb] 17

angle from l to m is equal to the angle from m to l ′. Clearly, the latter line is also the imageof l under the composition δ 2

ml . Therefore, equation (3.16.2) holds.(3) As in (2), the composition ρmρl is a Möbius transformation. Let n be the line through

P and Q, and let u and v be its limit points. As n is orthogonal to the lines l and m, thelimit points u and v are interchanged by any of the reflections ρm and ρl . Hence u and v arefixed points of the composition ρmρl . Therefore, the composition is a translation with axis n.Moreover, the point P is, by any of the two sides of equation (3.16.3), mapped to the point Ron n for which Q is the midpoint of the line segment PR. Therefore, the equation holds.

(3.17) Corollary. In the setup of (3.9), the following equations hold:

τ 2ACτ

2CBτ

2BA = 1, (3.17.1)

δ 2acδ

2cbδ

2ba = 1. (3.17.2)

In addition, consider the line h through A orthogonal to a, let H be the point of intersectionof h and a, and let a ′ be the line through A orthogonal to h. Then,

τ 2AH = δ 2

a′cδ2ca. (3.17.3)

Proof. Note that, depending on a choice of orientations of the sides a and c, there are tworotations δac. If one is by the angle θ , then the other is by the angle θ − π . Hence the squareδ 2ac is well defined.

Obviously, the first formula follows from (3.16.1) since the half turns are involutions.Similarly, the second formula follows from (3.16.2), and the third from (3.16.2) and (3.16.3).

(3.18) Lemma. Consider two matrices σ and τ in SL(D). Assume that τ is a translation inD, and assume that σ is either a rotation around a point on the axis of τ or a translationalong an axis orthogonal to the axis of τ . Then,

2 tr(στ) = tr(σ )tr(τ ). (3.18.1)

Proof. By conjugation we may assume that D is the upper half plane H, that the axis of τ isthe imaginary axis iR, and that either σ is a rotation around the point i or a translation alongthe line through i orthogonal to the imaginary axis.

Now τ and σ belong to SL2(R). The matrix τ fixes 0 and ∞; hence it is of the form,[a 00 d

]where ad = 1.

The matrix of σ is either a rotation or a translation. Accordingly, by (2.4), it is of one of thefollowing two forms, [

c b

−b c

]or

[c b

b c

],

where c2 + b2 = 1 or c2 − b2 = 1. Clearly, in both cases the left hand side of (3.18.1) isequal to 2c(a + d) and hence equal to the right hand side.

21


(3.19) Proposition. In the setup of (3.9), assume for the triangle ABC that the angle C is aright angle. Then,

cosh c = cosh a cosh b, cosB = cosh b sinA. (3.19.1)

Proof. Clearly, the two equations are equivalent to the equations of (3.10.1). To give analternative proof, note that the first equation of (3.17) implies the following:

τ 2AB = τ 2

ACτ2CB.

Moreover, as the angle at C is a right angle, Lemma (3.18) applies with σ := τ 2AC and

τ := τ 2CB . As a consequence, we obtain the equation,

2 tr(τ 2AB) = tr(τ 2

AC)tr(τ2CB).

Dividing by 4 we obtain an equation for the invariants h(σ ) which, by Lemma (3.14)(2) isthe first asserted equation (3.19.1).

To prove the second equation, consider the last equation in (3.17). As the angle at C is aright angle, we have that H = C. Thus we obtain the equation,

δ 2ca′τ 2

AC = δ 2ca.

Again Lemma (3.18) applies, and we obtain the equation,

2 tr(δ 2ca) = tr(δ 2

ca′)tr(τ 2AC).

Divide by 4 to obtain an equation for the invariants h(σ ). By construction of a ′, the anglefrom c to a′ is equal to π/2 −A. Hence, by Lemma (3.14)(1) and (2), the equation obtainedis the second equation of (3.19.1).

(3.20) Exercise. Consider two lines l and l ′ of D, with limit points u, v and u′, v′. Assumethat the four points u, v, u′, v′ on the boundary of D are different. In addition, assume thatv and v′ belong to the same of the two arcs in which the boundary ∂D is divided by u, u ′.Prove that the cross ratio df(u, v, u′, v′) is negative, if and only if the lines l and l ′ intersect.If the cross ratio is negative, then the angle θ between the lines l and l ′ is determined by theformula,

tan2(θ/2) = − df(u, v, u′, v′).

If the cross ratio is positive, then the (non-euclidean) distance θ between the lines is determinedby the formula,

tanh2(θ/2) = df(u, v, u′, v′).

22


[Möb] 19

4. Proper actions.

(4.1) Proposition. Let G := SL(D) be the stabilizer of a disk D in C. Then the action of Gon D is proper, that is, the map (α, w) �→ (αw,w) is a proper map,

G× D → D × D. (4.1.1)

Proof. Recall that a continuous map between locally compact (Hausdorff) spaces is a propermap if the preimage of any compact subset of the target is a compact subset of the source.

The map (4.1.1) is an obvious composition of two maps,

G× D → PG× D → D × D. (1)

Of the two maps, the first is proper, becauseG → PG is the homomorphism with kernel ±1.Thus is suffices to show that the second map is proper.

Fix a point u on the boundary ∂D. Then there is a map,

PG× D → ∂D × D × D,

defined by(α, w) �→ (αu, αw,w).

It is bijective by Corollary (1.9), and obviously continuous. It is in fact a homeomorphism, thatis, the inverse map is continuous. In other words, if we consider for (u′, w′, w) ∈ ∂D×D×Dthe unique Möbius transformation z �→ αz under which (D, w, u) is mapped to (D, w ′, u′),then the Möbius transformation depends continuously on (u ′, w′, w). The latter fact is easilyproved using the explicit determination of the Möbius transformation given in Section 1.

Clearly, when the source of the second map in (1) is replaced by the homeomorphic space∂D × D × D, then the map is simply the projection,

∂D × D × D → D × D.

Hence the map is proper, because ∂D is compact.Thus the Proposition has been proved.

(4.2) Corollary. LetK1 andK2 be compact subsets of the disk D. Then the following subsetof G := SL(D) is compact:

{α | K1 ∩ αK2 �= ∅}.

Proof. The product set K1 × K2 is a compact subset of D × D. Hence its preimage underthe map in (4.1) is a compact subset of G× D. Clearly, the preimage is the following set:

{(α, z) | αz ∈ K1 and z ∈ K2}.Hence the latter set is compact. Consequently, its image inGunder the projectionG×D → G

is a compact subset of G. The latter image is the set in question. Hence it is compact.

23


(4.3) Corollary. For any point z ∈ D, the map α �→ αz is a proper map,

G → D.

In particular, the isotropy groupGz is a compact subgroup of G.

Proof. The first assertion of the Corollary is the special case of (4.2) obtained by taking K1arbitrary compact and K2 = {z}. The second assertion is a consequence, since the isotropygroup is the preimage of the compact set {z} under the map G → D.

(4.4) Note. In Section 2 we proved that the isotropy group Gz is in fact conjugated to thecompact group SO2(R), and in particular, the isotropy group is isomorphic to the unit circleU1(C). Note also that we proved in Section 2 that the isotropy group SL(D)u of a point u onthe boundary ∂D is non-compact. In particular, the action of SL(D) on the closure of D isnot proper.

(4.5) Exercise. Several topological facts were used in the proof of Proposition (4.1). TheRiemann sphere C is the one point compactification of C. Hence the diskD as an open subsetof C is locally compact. The surjection of (1.1), (C2)∗ → C, is continuous and open. As aconsequence, C is equal to the quotient (C2)∗/C∗.

The group SL2(C) is locally compact: it is a closed subset of the space C4 of 2×2 matrices.It is a topological group. The group SL(D) is a closed subgroup, since it is conjugated toSL(H) = SL2(R).

The group PSL(D) has two topologies: the quotient topology induced by the surjectionSL(D) → PSL(D) and the subset topology induced by the inclusion PSL(D) ↪→ Autcont(D)(where the automorphism group is given the compact–open topology). The two topologiesare equal, and PSL(D) is a locally compact topological group.

Prove these topological facts. In addition, prove (or find a reference for) the followinggeneral facts: If K is compact and X is locally compact, then the projection K ×X → X isproper. If K is a compact subgroup of a locally compact group G, then the canonical mapG → G/K is proper.

24


[Discr] 1

Discrete subgroups

1. Isotropy groups and their generators.

(1.1) Setup. Fix a disk D and a discrete subgroup of SL(D). Recall that a subset of atopological space is said to be discrete in the given topological space, if it is a closed subsetand the induced topology is discrete. Equivalently, when the topological space is locallycompact, the condition is that the intersection of and any compact subset is finite. Inparticular, a subgroup of SL2(C) is discrete, if and only if, for every positive real numberR, there is only a finite number of matrices in for which all four entries have modulus atmost equal to R.

Recall that a point u is called -parabolic if it is a fixed point of some parabolic matrix in . The set of -parabolic points is denoted ∂ D. It is a subset of the boundary ∂D. A pointis -elliptic if it belongs to D and is fixed under some non-trivial (necessarily elliptic) matrixof . A -ordinary point is a point of D which is not -elliptic.

(1.2) Lemma. (1) Assume that is a discrete subgroup of SL(E). Then the isotropy group 0 of the point 0 in E is a finite cyclic group generated by the matrix,

d2π/N :=[e2πi/N 0

0 e−2πi/N

],

where N = | 0| is the order of the isotropy group. In particular, the point 0 is -elliptic,if and only if | 0| > 2. The group is homogeneous, if and only if N is even. The groupP 0 of associated Möbius transformations is cyclic, generated by the rotation z �→ e2πi/dz,where d = |P 0|.

(2) Assume that is a discrete subgroup of SL(H). Assume that the point ∞ on theboundary of H is -parabolic. Then the isotropy group ∞ is infinite, and all its matricesdifferent from ±1 are parabolic. The isotropy group it is either cyclic or dicyclic (that is,isomorphic to {±1} × Z). The group is homogeneous, if and only if the isotropy group ∞is dicyclic. In the dicyclic case, the isotropy group consists of all matrices of the form ±t nh(for a unique h > 0) where th is the matrix,

th :=[ 1 h

0 1

].

In the cyclic case, ∞ consists either of all powers t nh [the ‘regular’ case] or of all powers(−th)n [the ‘irregular’ case] (for a unique h > 0). In all cases, the group P ∞ of associatedMöbius transformations is the infinite cyclic group generated by the translation z �→ z+ h.

25

[Discr] 2 Automorphic functions26. februar 1995

Proof. Assertion (1) follows immediately from Lemma (Möb.2.3)(2) since 0 is a discretesubgroup of SL(E)0, and SL(E)0 is isomorphic to the unit circle U1(C).

To prove Assertion (2), note that the matrices of ∞ are of the form (Möb.2.3.1). Therefore,since the point ∞ is assumed to be -parabolic, there exists a matrix of the form ±th in .Moreover, since is discrete, we can choose in ∞ a matrix ±th with h > 0 minimal.Then, again using that is discrete, it follows from the relation (Möb.2.3.2) that no matrix(Möb.2.3.1) with a �= ±1 can belong to ∞. Thus all matrices different from ±1 in ∞ areparabolic. Again, since is discrete and h is minimal, it follows that all matrices of ∞ are,up to multiplication by ±1, powers of th. The remaining assertions of (2) follow easily.

(1.3) Definition. In (1), the generator z �→ e2πi/dz of P 0 is the Möbius transformationassociated to the following matrix:

γ0 :=[−eπi/d 0

0 −e−πi/d]. (1.3.1)

The matrix γ0 belongs to 0. Indeed, assume first that N is even. Then −1 ∈ 0 andd = N/2. As γ0 = −d2πi/N , it follows that γ0 ∈ 0. Assume next that N is odd. Then

d = N . As γ0 = d(N+1)/22πi/N , it follows again that γ0 ∈ 0.

The matrix γ0 is called the canonical generator at the point 0. It generates the group 0if N is odd or if N ≡ 0 (mod 4). If N ≡ 2 (mod 4), then γ0 generates an inhomogeneoussubgroup of index 2 in 0.

case, and the matrix γ∞ := th in the other cases, will be called the canonical generator atthe point ∞. The transformation associated to the canonical generator is map z �→ z+h, andh is the minimal possible step length of the transformations in P ∞. The canonical generatorγ∞ generates a cyclic inhomogeneous subgroup of ∞ that is mapped isomorphically ontoP ∞.

(1.4) Definition. For a general disk D and a point u in D ∪ ∂ D we obtain, by conjugation,assertions corresponding to those of Lemma (1.2).

If u is in D, choose a conjugation α : D → E such that αu = 0. Then is conjugate to thediscrete subgroup α α−1 of SL(E), and u is conjugate to 0. Hence the isotropy group u isconjugate to the isotropy group (α α−1)0. It follows from (1.2)(1) that u is a finite cyclicgroup, and hence its quotient P u is a finite cyclic group. The quotient P u is non-trivial ifand only if u is -elliptic. In any case, the order eu := |P u| is called the order of u withrespect to . The isotropy groups for the points in the orbit u are conjugate. In particular,all points in the orbit u have the same order. Moreover, if u is -elliptic, then all points inthe orbit u are -elliptic. In this case, the orbit is said to be a -elliptic orbit.

If u is in ∂ D, choose a conjugation α : D → H such that αu = ∞. It follows similarlythat the isotropy group u is infinite cyclic or dicyclic, and that its quotient P u is an infinitecyclic group. Moreover, all non-trivial matrices in u are parabolic. Clearly, all points in theorbit u are -parabolic. The orbit is said to be a -parabolic orbit or to be a cusp for .

Foru ∈ D∪∂ D, define the canonical generator of atu as the following matrix γu ∈ u:If u is in D, choose a conjugation α : (D, u) → (E, 0). Then is conjugate to the discrete

26


[Discr] 3

subgroupα α−1 of SL(E), andu is conjugate to 0. Define γu as the conjugate of the canonicalgenerator γ0 in (1.3). Similarly, if u is in ∂ D, choose a conjugation α : (D, u) → (H,∞),and define γu as the conjugate of γ∞.

In both cases, it has to be verified that the matrix obtained is independent of the choice ofα. However, two different choices α1 and α2 differ by a matrix α := α1α

−12 that stabilizes

the image disk and the image point α1u = α2u. We may assume that α has determinant 1.Thus is has to be proved, in the two cases (D, u) = (E, 0) and (D, u) = (H,∞) respectively,that if α ∈ SL(D)u, then, for the canonical generators considered in (1.3), conjugation by αmaps the canonical generator of (α α−1)u onto the canonical generator of u.

The latter assertion is obvious for (E, 0), since the isotropy SL(E)0 is commutative andhence conjugation byα is in fact the identity. For (H,∞), the assertion follows from Equation(Möb.2.3.2) describing conjugation of a canonical generator under a matrix in SL(H)∞.

In the parabolic case, the cusp represented by u is said to be a regular cusp if the canonicalgenerator γu is conjugate to th and to be an irregular cusp if γu is conjugate to −th. Note thatthe width h of th is not canonical. A different choice of conjugation changes th to ta2h, see(Möb.2.3.2).

(1.5) Example. The Cayley transformation z �→ (z − i)/(z + i) of (Möb.1.5) is associatedto the matrix,

α := 1

1 + i

[ 1 −i1 i

].

The Cayley transformation maps (H, i) onto (E, 0). Consequently, under conjugation by α,the stabilizer group SL(E) = SU1,1(C) is mapped onto the stabilizer group SL(H) = SL2(R),and the isotropy group SL(E)0 = U1(C) is mapped onto the isotropy group SL(H)i =SO2(R). Under the conjugation, the following matrices correspond:

[eiθ 00 e−iθ

]and

[ cos θ sin θ− sin θ cos θ

].

Consider the discrete subgroup := SL2(Z) of SL(H). It is homogeneous. Clearly, thepoint ∞ is -parabolic, and the canonical generator at ∞ is the matrix,

t :=[ 1 1

0 1

]. (1.5.1)

The point i = e2πi/4 is elliptic: the isotropy group i is of order 4 and the canonical generatorat i is the matrix of order 4,

s :=[ 0 −1

1 0

]. (1.5.2)

Indeed, the isotropy group i consists of the matrices with integer entries in SO2(R), andclearly, the latter matrices are the four matrices ±1, ±s. It follows from the correspondenceabove that −s is conjugate to the matrix d2π/4. Hence s is the canonical generator.

27


The point ρ = e2πi/3 is elliptic: the isotropy group ρ is of order 6 and the canonicalgenerator at ρ is the matrix of order 3,

u = t−1s =[−1 −1

1 0

]. (1.5.3)

Indeed, since ρ2 + ρ + 1 = 0, it is easy to describe the isotropy group SL(H)ρ . It followseasily that the isotropy group ρ consists of the following 6 matrices ±1, ±u, ±u2. Thematrix u is of order 3. Hence ρ is generated by the two matrices −u and −u2 of order 6.The Möbius transformation associated to ±u is the rotation by the angle 2π/3 since the halfray from ρ to ∞ is mapped to the half ray from ρ to −1. Hence −u is conjugate to the matrixd2π/6, and u is the canonical generator.

(1.6) Proposition. Assume that is the discrete subgroup SL2(Z) of SL(H). Let F be theclosed subset of H defined by the inequalities,

F : |z| ≥ 1, −1

2≤ z ≤ 1

2.

Denote by F0 the subset obtained from F by omitting from the boundary the points z whereeither z = 1/2 or |z| = 1, z > 0. Then the set F0 is a complete set of representativesfor the set of orbits H/ . Moreover, the points i and ρ are the only -elliptic points inF0. Finally, there is only one cusp, namely the orbit ∞ consisting of ∞ and the rationalnumbers.

Proof. In addition to the subsets F and F0, denote byG the vertical strip of H defined by theinequalities: − 1

2 ≤ z < 12 . We proceed in a series of steps.

Step 1. For any point z of H there is only a finite number of matrices σ in ,

σ =[a b

c d

], (1)

such that(σz) ≥ z, σz ∈ G. (2)

Indeed, for a fixed pair of integers (c, d), consider the matrices (1). They correspond to theinteger solutions (a, b) of the equation ad − bc = 1. Hence, for the fixed pair of primeintegers (c, d), there are matrices σ of the form (1), and if σ0 is any such matrix, then thematrices of the form (1) are exactly the matrices t kσ0 for k ∈ Z. The matrix t defines thetranslation tz = z + 1 and the vertical strip G is of width 1. Therefore, when (c, d) is fixed,there is only one matrix σ of the form (1) such that σz ∈ G. Now, for any matrix (1) inSL2(R), we have that (σz) = (z)/|cz + d|2. Hence the inequality in (2) is equivalent tothe following:

|cz + d|2 ≤ 1. (3)

The latter inequality has only a finite number of integer solutions (c, d). Therefore, theassertion of Step 1 holds.

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[Discr] 5

Step 2. If |z| < 1, then (sz) > z. This is just a simple observation.Step 3. Any point z in H is -equivalent to a point in F0. Indeed, for any point z there is a

unique point z1 in G of the form t kz. If |z1| < 1, consider z2 := (sz1)1. Define, inductively,zn+1 := (szn)1 as long as |zn| < 1. By construction, the points zj belong to G, and theyare -equivalent to z. By Step 2 we have that z1 < z2 < · · · . Therefore, by Step 1, thesequence z1, z2, . . . is finite. It stops with a point zn in G such that |zn| ≥ 1. If |zn| > 1,then zn belongs to F0. If |zn| = 1, then either zn or szn belongs to F0. Hence, in all cases,the point z is -equivalent to a point of F0.

Step 4. By Step 3, the setF0 contains a complete set of representatives forH/ . Therefore,to show that F0 is a complete set of representatives and that i and ρ are the only -ellipticpoints in F0, it suffices to prove the following assertion: Let σ �= ±1 be a matrix in , andlet w and w′ be points of F0. Then the equation σw = w′ implies that

(s) either w′ = w = i and σ = ±s,(u) or w′ = w = ρ and σ = ±u,±u2.

To prove the assertion, we may assume that w′ ≥ w. Then 1 ≥ |cw+ d|2. Hence the firstof the following four inequalities holds:

1 ≥ c2|w|2 + cd(w+w)+d2 ≥ c2 + cd(w+w)+d2 ≥ c2 −|cd|+d2 ≥ (|c|− |d|)2. (4)

The second inequality holds because |w| ≥ 1. The third holds because | w| ≤ 12 , and the

last is trivial. Moreover, the last inequality is strict if cd �= 0.Since c and d are integers, the two last expressions in (4) are non-negative integers.

Moreover c and d are prime. Hence the inequalities leave the three possibilities: c =0, |d| = 1, or |c| = 1, d = 0, or |c| = |d| = 1.

Assume first that c = 0, |d| = 1. Replacing σ by −σ we may assume that d = 1. Thenthe matrix σ is of the form,

σ =[ 1 b

0 1

],

and w′ = σw = w + b. As w′ and w belong to F0 by hypothesis, it follows that b = 0,contradicting that σ �= ±1. Thus the first possibility is excluded.

Assume next that |c| = 1, d = 0. We may assume that c = 1. Then the matrix σ is of theform,

σ =[a −11 0

],

andw′ = σw = sw+a. Clearly, in (4) the second inequality is an equality, and consequently|w| = 1. Asw′ andw belong toF0 by hypothesis, it follows that either a = 0 andw = w ′ = i,or a = −1 and w′ = w = ρ. In the first case, σ = s and hence (s) holds, in the second case,σ = u and hence (u) holds.

Finally, assume that |c| = |d| = 1. We may assume that d = 1 and c = ±1. Thenthe fourth inequality in (4) is strict. Consequently, the first three inequalities are equalities.Now, the second inequality is strict unless |w| = 1 and the third inequality is strict unless

29


w = − 12 and cd = 1. Therefore, it follows that w = ρ and that c = d = 1, that is, σ has

the formσ =

[a a − 11 1

].

Then w′ = σw = (−1)/(w + 1) + a = ρ + a. As w′ is in F0 by hypothesis, it followsthat a = 0. Thus w′ = w = ρ and σ = −u2. Hence (u) holds. Thus the proof of Step 4 iscompleted.

Step 5. To prove the final assertion of the Proposition, note that the point ∞ is -parabolic.Clearly, the orbit ∞ is equal to Q := Q ∪ {∞}. Conversely, let u ∈ R be a -parabolicpoint. Then u is a fixed point of a non-trivial matrix γ in with eigenvalue ±1. ThereforeJ (γ, u) = ±1. Hence u ∈ Q, because J (σ, u) has the form cu+ d, with c, d ∈ Z and c �= 0.

Thus all assertion of the Proposition have been proved.

(1.7) Remark. The group = SL2(Z) is the modular group. It is generated by the matricess and t . Indeed, let 1 be the subgroup generated by s and t , and fix a point w in the interiorof F0. Let γ be a matrix of . It follows from the proof of (1.6) applied to z := γw thatthere is a matrix γ1 in 1 such that γ1γw belongs to F0. As w and γ1γw both belong to F0,it follows that γ1γ = ±1. Hence γ = ±γ−1

1 belongs to 1.

(1.8) Lemma. Let� be a subgroup of finite index in . Then every -orbit splits into at mostthe number | :�| of �-orbits. Moreover, every �-elliptic point is -elliptic, and a point is�-parabolic if and only if it is -parabolic. Finally, the canonical generator of � at a pointu is up to a sign equal to γ duu , where du is the index, du := |P u:P�u|.Proof. Set d := | :�|, and let γj be a system of representatives for the right cosets modulo�.Consider a point u of D∪ ∂D. Then the orbit u is the union, u = �(γ1u)∪ · · · ∪�(γdu)of the number d of�-orbits. Of these, at most d are different. Hence the first assertion holds.

Clearly, a�-elliptic (resp.�-parabolic) point u is -elliptic (resp. -parabolic). To provethat the converse holds for a -parabolic point u, note that the isotropy group �u is of finiteindex (at most d) in u. Hence�u is infinite, because u is infinite. Moreover, all nontrivialmatrices in u are parabolic. Therefore �u contains a parabolic matrix.

The final assertion is left to the reader.

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[Discr] 7

2. Properly discontinuous actions.

(2.1) Setup. Consider, as in Section 1, a disk D and a discrete subgroup of SL(D).

(2.2) Lemma. Let K1 and K2 be compact subsets of the disk D. Then there is only a finitenumber of elements γ in for which the following condition holds:

K1 ∩ γK2 �= ∅.

Proof. The set of elements γ for which the condition holds is the intersection of and thesubset of SL(D) described in (Möb.4.2). As the latter set is compact, and is discrete, theintersection is finite.

(2.3) Remark. An action of a group on a locally compact space is said to be properlydiscontinuous if it has the property of Lemma (2.2).

(2.4) Corollary. For every point u of D, the stabilizer u is a finite group and the orbit uis discrete in D.

Proof. For every compact subsetK ofD, the condition γ u ∈ K holds for only a finite numberof elements γ in . In particular, the intersection u ∩ K is finite. Therefore the orbit uis discrete in D. The first assertion of the Corollary was proved in Section 1; alternatively, itfollows from (2.2) by taking K1 = K2 := {u}.(2.5) Corollary. For any two points u and u′ in D, there are neighborhoods U of u and U ′of u′, such that for any matrix γ in , if γU ∩ U ′ �= ∅, then γ u = u′. In particular, everypoint u of D has a neighborhood U such that for any matrix γ in , if γU ∩ U �= ∅, thenγ ∈ u.

Proof. Choose open neighborhoods V of u and V ′ of u′ whose closures relative to D arecompact. Obviously, the condition,

γV ∩ V ′ �= ∅, (1)

holds for the matrices γ in for which γ u = u′. By Proposition (2.2), applied to theclosures of V and V ′, the condition (1) can hold only for a finite number of elements of . Inparticular, there is only a finite number of elements γ1, . . . , γN in such that (1) holds andγiu �= u′. For each i = 1, . . . , N , the point γiu is different from u′. Accordingly, there areopen neighborhoods Ui of u and U ′

i of u′ such that γiUi ∩ U ′i = ∅. Now, clearly, the two

open sets,U := V∩U1∩ · · · ∩UN, U ′ := V ′∩U ′

1∩ · · · ∩U ′N,

have the required properties.When u′ = u, the asserted special case is obtained by replacing U by U ∩ U ′.

31


(2.6) Corollary. The set of -elliptic points is a discrete subset of D.

Proof. Otherwise there would be a point u in D which is an accumulation point of the setof -elliptic points. Then every neighborhood U of u contains an infinity of -ellipticpoints ui . For each ui there is an elliptic transformation γi in with ui as fixed point. Thetransformations γi are different, because every elliptic transformation has only one fixed pointin D. In particular, there is an infinite number of elements γi in for which γiU ∩ U �= ∅.Thus we have obtained a contradiction to the assertion in (2.5).

(2.7) Lemma. Consider the group G = SL2(R) acting on the upper half plane H. Letll :G → R be the map defined by [

a b

c d

]�→ c .

Let� be a discrete subgroup ofG. Assume that the point ∞ is�-parabolic. Then the imagell(�) is discrete in R.

Proof. Since the point ∞ is �-parabolic, there is in � a matrix of the form,

τ =[ 1 h

0 1

]where h > 0.

We have to prove, for any given positive real number R, that the intersection ll(�) and[−R,R] is finite.

For 0 < ε < 1, denote by Kε the set of all matrices δ in G for which the followinginequalities hold:

ε ≤ (δi) ≤ 1, 0 ≤ (δi) ≤ h. (1)

The setKε is a compact subset ofG. Indeed, the inequalities above, for the real and imaginaryparts of complex numbers, describe a compact rectangle in H, and the setKε is the preimageof the rectangle under the map δ �→ δi. As the map δ �→ δi is proper by Corollary (Möb.4.3),the preimage is compact.

The intersection�∩Kε is finite, because� is discrete inG andKε is compact. Therefore,to prove that the intersection ll(�) ∩ [−R,R] is finite, it suffices to prove that when ε ischosen sufficiently small, then the intersection ll(�) ∩ [−R,R] is contained in the imagell(�∩Kε). We prove that the latter inclusion holds when ε ≤ 1/

(R2 + (1 + hR)2

).

Let c be a value in the intersection ll(�)∩ [−R,R]. If c = 0, then c = ll(1), and clearlythe identity matrix 1 belongs to�∩Kε for any ε. Assume that c �= 0. Then 0 < |c| ≤ R andthere exists in � a matrix,

δ =[a b

c d

], (2)

whose lower left entry is the given value c. If δ is replaced by δτ , the c is unchanged and thed is replaced by d + hc. Therefore, by replacing δ by a product δτ n for a suitable n, we mayassume in (2) that 1 ≤ d ≤ 1 + h|c|. Then,

1 ≤ c2 + d2 ≤ 1/ε. (3)

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[Discr] 9

Indeed, the first inequality holds because 1 ≤ d, and the second holds by the choice of ε sinced ≤ 1 + h|c|.

As the matrix δ has determinant 1, the imaginary part (δi) is equal to 1/(c2 +d2). Hencethe first two inequalities of (1) hold by (3). When δ is replaced by τδ, then the c and the dare unchanged, but the value δi is replaced by δi + h. Therefore, by replacing δ by τmδ fora suitable m, we may assume that the last two inequalities of (1) hold. Hence δ belongs to� ∩Kε and, consequently, c belongs to ll(�∩Kε).

Thus the required inclusion has been proved.

(2.8) Definition. Let u be a point on the boundary ∂D. Then, by definition, a fundamentalneighborhood of u is the union of the point u and an open disk (strictly) contained in D andtangent to ∂D at the point u. The boundary of the disk, excluding the point u, is called a horocycle.

Clearly, the topology on D is the trace of a topology on the closed disk,

D ∪ ∂D,

in which a neighborhood of a point u of ∂D is a subset of D∪ ∂D containing a fundamentalneighborhood of u. By convenience, a fundamental neighborhood of a point u in D is ageodesic disk around u, that is, an open ball centered at u with respect to the non-euclideandistance on D.

Clearly, a Möbius transformation z �→ αzmaps a fundamental neighborhoodU of a pointu of D ∪ ∂D onto a fundamental neighborhood of the image point αu in αD ∪ ∂αD. Inparticular, any matrix in SL(D) defines a topological automorphism of D ∪ ∂D.

Clearly, the subset ∂ D of -parabolic points is invariant under the action of . Hence acts on the topological space D ∪ ∂ D. We denote by D/ the corresponding orbit space,with its quotient topology. Note that D/ contains the quotient D/ as an open subspace.In addition, it contains one point for each orbit of -parabolic points. A point in D/ iscalled parabolic, elliptic, or ordinary respectively, if it corresponds to an orbit of -parabolicpoints, an orbit of -elliptic points, or an orbit of -ordinary points.

(2.9) Example. In the topology of the closed disk H ∪ R, the fundamental neighborhoodsof the point ∞ on the boundary are the subsets containing ∞ and a half plane HR : z > R

for some positive real number R. The horo cycles around ∞ are the horizontal straight linescontained in H.

(2.10) Lemma. Let K be a compact subset of D. Then, for every -parabolic point u, thereexists a neighborhood U of u such that for all matrices γ ∈ ,

γU ∩K = ∅.

Proof. After a conjugation, we may assume that D is the upper half plane H and that u = ∞.From Lemma (2.7), it follows in particular that the exists a positive real number r with the

33


property that if γ is a matrix in and |ll(γ )| < r , then ll(γ ) = 0. As K is a compact subsetof H, it is contained in a horizontal strip,

K ⊆ {z ∈ H | r1 < z < r2} (1)

Now choose R such that R ≥ r2 and R ≥ 1/(r1r2). LetU be the corresponding fundamentalneighborhood, HR : z > R, of ∞. We have to prove for any point z of U and any matrix γin ,

γ =[a b

c d

],

that γ z /∈ K .Now, since R ≥ r2, the half plane U is disjoint from the strip in (1), and in particular

disjoint fromK . Assume first that |c| < r . Then c = ll(γ ) = 0, and hence the transformationz �→ γ z is of the form z �→ z+ h. Consequently, if z ∈ U , then γ z ∈ U , and it follows thatγ z /∈ K . Assume next that |c| ≥ r . Then,

(γ z) = 1

|cz + d|2 z ≤ 1

(|c|z)2 z ≤ 1

r2z .

If z ∈ U , then the right hand side is less than 1/(r 2R), and hence, by the choice of R, theright hand side is less than r1. Consequently, γ z /∈ K .

(2.11) Lemma. Consider an orbitB = w0, wherew0 is a point of D. IfU is a fundamentalneighborhood of a point u in D, then U contains only a finite number of points in B. If U isa fundamental neighborhood of a point in ∂ D, then of the horo cycles contained in U , onlya finite number contain points of B.

Proof. The orbitB is discrete by (2.4). Hence the assertion for a pointu ∈ D follows, becausea fundamental neighborhood U has compact closure.

Assume that u is -parabolic, and let U be any fundamental neighborhood u. Afterconjugation, we may assume that (D, u) = (H,∞). Then U is a half plane HR : z > R,and the horo cycles are the straight horizontal lines. By (2.10) applied with K = {w0}, thehalf plane HR for R � 0 contains no points of B. Consider the canonical generator γ∞ of ∞. It is a translation z �→ z+h. Then every point in B ∩Hr is mapped by a suitable powerof γ∞ in the rectangle of H:

0 ≤ z ≤ h, r ≤ z ≤ R.

The rectangle is compact. Hence, by (2.4), it contains only a finite number of points of B.Thus every point of B ∩ U is one of the horo cycles through these finitely many points.

(2.12) Theorem. For any two points u1 and u2 in D ∪ ∂ D, there are neighborhoods U1of u1 and U2 of u2, such that for any matrix γ in , if γU1 ∩ U2 �= ∅, then γ u1 = u2. Inparticular, every point u of D∪ ∂ D has a neighborhoodU such that for any matrix γ in ,if γU ∩ U �= ∅, then γ ∈ u.

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[Discr] 11

Proof. If u1 and u2 belong to D, the assertion is the content of Corollary (2.5). If u1 is -parabolic and u2 ∈ D, the assertion follows from Lemma (2.10) by taking asU2 any compactneighborhood of u2.

Assume that u1 and u2 are -parabolic. After a conjugation, we may assume that D is theupper half plane H and that u2 = ∞. Let ±τh be the canonical generator of ∞. Choose inH any straight horizontal line L, and on L any horizontal line segment K of length at leastequal to h. Then K is a compact subset of H, and so Lemma (2.10) applies with u := u1.Accordingly, there is a fundamental neighborhood U1 of u1 such that γU1 ∩K = ∅ for all γin . Choose as neighborhood U2 of u2 = ∞ any fundamental neighborhood HR : z > R

lying above the line L, that is, such that R is at least equal to the imaginary part of points onL.

We claim that if γU1 ∩ U2 �= ∅, then γ u1 = ∞. Indeed, the image γ u1 is a point on theboundary of H. Assume that γ u1 �= ∞. Then γ u1 is a point on the real axis, and γU1 is a(usual) disk in H tangent to the real axis. By assumption, the disk contains a point inU2, thatis, a point with imaginary part greater thanR. Therefore, the disk γU1 contains a point on theline L. A suitable power τ nh will move the latter point into the set K . Thus τ nh γU1 ∩K �= ∅,contradicting the property of U1.

Thus the Theorem has been proved.

(2.13) Corollary. The quotient D/ is a Hausdorff space. Moreover, it is locally compact.In fact, for any point u inD∪∂ D, ifU is a sufficiently small fundamental neighborhood of u,then every point different from u inU is ordinary and the image ofU in D/ is topologicallyisomorphic to an open disk: {q ∈ C : |q| < ε}. In particular, the sets of -parabolic pointsand -elliptic points are discrete in D/ .

Proof. It follows from Theorem (2.12) that the topology on the quotient space D/ is Haus-dorff. Indeed, let u1 and u2 represent two given different points of the quotient space. Thenu1 and u2 are not -equivalent. Therefore, by (2.12) there are open neighborhoods U1 of u1and U2 of u2 such that γU1 ∩U2 = ∅ for all γ in . Consequently, the images of U1 and U2in the quotient are disjoint. As the quotient map is an open map, the two images are disjointneighborhoods of the two given points.

Consider next a given point in the quotient space, represented by a point u. By (2.12), fora sufficiently small fundamental neighborhood U of u, the following condition holds:

γU ∩ U �= ∅ �⇒ γ ∈ u. (1)

Except possibly for u, all the points of U are in D, and hence not parabolic. Moreover, nopointw different from u inU can be elliptic. Indeed, ifw inU is elliptic, sayw = γw whereγ �= ±1, then it follows from (1) that also u is a fixed point of γ , and then w = u because anelliptic Möbius transformation has its other fixed point in the disc exterior to D. Hence allpoints different from u in U are ordinary.

As the quotient map is an open map, the image of U in D/ is an open neighborhood ofthe given point, and topology on the image as a subset is equal to the quotient topology onthe image. It follows from (1) that two points in U are equivalent under if and only if they

35


are equivalent under u. Moreover, since U is a fundamental neighborhood of u, it is stableunder u. Therefore, the image of U in D/ is isomorphic to the quotient U/ u.

Assume first thatu is -parabolic. After conjugation we may assume that (D, u) is (H,∞).Then U , as a fundamental neighborhood of ∞, is a half plane z > R. Moreover, the groupP ∞ is the infinite cyclic group generated by z �→ z+ h. Consider the exponential,

q(z) := e2πiz/h.

It defines a continuous map q from H to the pointed unit disk {w : 0 < |w| < 1}, and theinduced equivalence relation on H is precisely the equivalence relation defined by the actionof ∞ on H. Moreover, the map q defines a continuous open map from U onto an open disk{w : |w| < ε}. Therefore q induces a topological isomorphism,

U/ ∞ ∼−→{w : |w| < ε}.

Assume next thatu is a point inD. After conjugation, we may assume that (D, u) = (E, 0).Then U , as a fundamental neighborhood of 0 in the unit disk, is an ordinary disk: |z| < ε.Moreover, the group P 0 is a finite cyclic group of order d, generated by the map z �→ ζ z

for some d-th root of unity ζ . Consider the d’th power map,

q(z) = zd.

It defines a continuous map q from E onto itself, and the induced equivalence relation on Eis precisely the equivalence relation defined by the action of 0 on E. Moreover, the map qdefines a continuous open map from the diskU onto itself. Therefore q induces a topologicalisomorphism,

U/ 0∼−→{w : |w| < ε}.

Hence we have proved in both cases that the image of U in D/ is isomorphic to an opendisk as asserted.

Hence all assertions of the Corollary have been proved.

(2.14) Corollary. If the quotient D/ is compact, then the numbers of -parabolic pointsand -elliptic points in D/ are finite. If the quotient D/ is compact, then there are no -parabolic points.

Proof. Clearly, the first assertion follows from Corollary (2.13). The quotient D/ is anopen subset of D/ , and it is dense by (2.13). Therefore, the second assertion holds.

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[Discr] 13

3. Finite normal fundamental domains.

(3.1) Setup. Fix a disk D and a discrete subgroup of SL(D). A subset F of D is called afundamental domain for the action of on D if the following three conditions are satisfied:

(1) The transforms γF for γ in cover D.(2) The set F is the closure in D of its subset U of interior points.(3) For any γ �= ±1 in , the transform γU is disjoint from U .

It follows from (2) that a fundamental domain F is a closed subset of D and that anyneighborhood of a point in the boundary F − U contains points from U and points from thecomplement of F . It follows from (1) that F contains a system of representatives for the orbitspace D/ . It follows from (3) that if an orbit has a representative in the interior U , then ithas no other representatives in F . However, some orbits might have several representativesbelonging to the boundary F −U . The orbit space D/ is equal to the quotient of F modulothe identification of -equivalent points on the boundary.

We will show in the next section that any discrete subgroup of SL(D) has a fundamentaldomain. In this section we fix a fundamental domainF , and we consider additional conditionson F . At a minimum we assume the following condition:

(4) The boundary F − U is the union of a finite number of line segments.

(3.2) Definition. The line segments forming the boundary of F will be called the boundarysegments. Note that a boundary segment can be infinite: one or both of its end points may belimit points, that is, they may belong to the boundary ∂D. If two of the boundary segmentslie on the same line and have points in common, we may replace the two by their union, andconversely, any boundary segment can be divided into two by an interior point. Thus we mayassume that if two of the boundary segments meet, then they meet at a common end point.The end points of the boundary segments are called the vertices of F . The finite vertices,that is, the vertices in D, belong to F , the infinite vertices are limit points of F . An infinitevertex which is isolated among the limit points of F is said to be a cusp of F . The domainF is called a finite domain if all of its infinite vertices are cusps. Equivalently, F is finite if ithas only a finite number of limit points.

(3.3) Observation. Take any point v ofF , and a sufficiently small fundamental neighborhoodV of v. If v is an interior point of F , then V is contained in U . If v is a boundary point of F ,and not a vertex, then there is a boundary segment through v such that F ∩ V is one of thetwo sectors into which V is divided by the line determined by the boundary segment. Finally,if v is a vertex, then F ∩ V consists of one or more angular sectors of V , each bounded twohalf rays through v. In the latter case, the sum of the angles of the angular sectors is calledthe angle of F at v. In the two former cases, the angle at v is, respectively, 2π and π .

Take similarly a limit point v of F and a sufficiently small fundamental neighborhood Vof v. If v is not a vertex, then (except for v) all point of V are contained in U . If v is a cusp,then F ∩V is a union of finitely many cuspidal sectors each bounded by two half rays throughv. Finally, if v is a vertex and not a cusp, then F ∩ V consists of a finite number (possiblynone) of cuspidal sectors and one or two sectors bounded by a single half ray through v.

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It follows from Condition (3) that if a point of F is -elliptic, then it lies on the boundaryof F . Moreover, if the order of the point is at least 3, then it is even a vertex of F . Similarly,if a limit point of F is -parabolic, then it is a cusp of F .

(3.4) Definition. A line segment L in D is called a side of F if it is the intersection of aboundary segment of F and a boundary segment of γF for some γ �= ±1 in . The pointsof L are then common boundary points of F and γF , and each end point of L is a vertex ofeither F or γF . In particular, an end point of a side is -equivalent to a vertex of F . Clearly,the transformation γ is unique. The inverse γ −1 is called the boundary transformationcorresponding to the side L, and it is denoted γL. It maps the side L onto a side γLL of F .

By Condition (3), the intersection F ∩ γF is contained in the boundary of F and in theboundary of γF . Each of the two boundaries is a finite union of boundary segments. Itfollows that the intersection is a union of a finite number of points and a finite number ofsides of F .

(3.5) Example. Consider the subset F = F(1) of H introduced in Proposition (1.6). Itfollows from the Proposition that F is a finite fundamental domain for the action of SL2(Z)on H. The boundary of F is the union of 3 line segments, namely the segment from ∞to ρ, the segment from ρ to ρ + 1, and the segment from ρ + 1 to ∞. The finite verticesare ρ and ρ + 1, and ∞ is the infinite vertex. The angles F at ρ and ρ + 1 are equal to2π/6. Moreover, the vertex ∞ is a cusp of F . Clearly, the three boundary segments are sidesof F . The corresponding boundary transformations are t : z �→ z + 1, s : z �→ −1/z, andt−1 : z �→ z − 1.

(3.6) Proposition. For any point w of D there is only a finite number of transforms γF suchthat w ∈ γF . In particular, in F there is only a finite number of points that are -equivalentto w. Moreover, F has only a finite number of sides.

Proof. Assume that w belongs to the n transforms γ1F, . . . , γnF , and consider the union,

γ1F ∪ · · · ∪ γnF. (3.6.1)

Each transform γiF is, in a sufficiently small fundamental neighborhood of w, a union ofangular sectors, and the angle of γiF at w is equal to the angle of F at γ −1

i w. The interiorsof different transforms are disjoint. Hence, in a sufficiently small fundamental neighborhoodof w, the union (3.6.1) is a union of angular sectors. The angle of the union is at most 2π andit is equal to the sum of the angles of F at the points γ −1

i w. The latter angles are boundedaway from 0, since the angle is at least equal to π except possibly at the finitely many verticesof F . Therefore, there is an upper limit for the number n. Thus there is only a finite numberof transforms γjF containing w.

If v ∈ F is -equivalent to w, say w = γ v, then γF is among the γjF , and hence thetransformation γ is equal to one of the γj . In particular, v is one of the finitely many pointsγ−1j w. Hence, only a finite number of points of F are equivalent to w.

An end point v of a side of F is -equivalent to a vertex of F . If v is a limit point of F ,then v is an infinite vertex of F . If v is in F , then v is -equivalent to a finite vertex of F and

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so, by what was just proved, there is only a finite number of possibilities for v. Hence thereis only a finite number of end points of sides. Thus the number of sides of F is finite.

(3.7) Observation. Consider the union (3.6.1) with n maximal, that is, the γjF are all thetransforms containing w. Then, clearly, the union is a neighborhood of w if and only if thesum of the angles of F at the points γ −1

j w is equal to 2π .

(3.8) Note. It follows from Proposition (3.6) that -equivalence on F is finite, that is, everypoint v of F is -equivalent to only a finite number of points in F . If v is in U , then v is -equivalent only to itself. If v is a relative interior point of a sideL of F , then v is equivalentto γLv and to no other point different from v.

It follows also that F contains only a finite number of -elliptic points v. Indeed, if vis -elliptic of order at least 3, then v is one of the finitely many vertices. So assume thatv is -elliptic of order 2, and not a vertex, that is, assume that v is relative interior on aboundary segment B. The non-trivial transformation γ with v as fixed point is a rotation byπ around u. Therefore B has a line segment in common with γF . It follows that v belongsto a side of F and that γ is the corresponding boundary transformation. There is only a finitenumber of sides. In particular, only a finite number of points can be fixed point of a boundarytransformation. Hence there is only a finite number of possibilities for v.

(3.9) Proposition. The following conditions on the domain F are equivalent:

(5i) Any compact subset K of D meets only a finite number of transforms γF .(5ii) For any point w of D, the union,

γ1F ∪ · · · ∪ γnF, (3.9.1)

where the union is over the finitely many transforms γjF containing w, is a neigh-borhood of w.

(5iii) The boundary of F is the union of the sides of F .

Proof. (i) �⇒ (ii): Let W be a fundamental neighborhood of w. The transforms γF coverD. In particular, they cover W . As the closure of W is compact, it follows from (i) that Wis contained in a finite union of transforms γjF . Each transform is a closed subset of D.Therefore, if we omit from the transforms γjF those that do no contain w, then the union ofthe remaining γjF is still a neighborhood of w. Thus (ii) holds.

(ii) �⇒ (i): For any point w of D, the union (3.9.1) of the finitely many γjF containsa fundamental neighborhood W of w. Each transform γF is the closure of its interior.Therefore, if a transform γF meets W , then it is equal to one of the γjF . In particular,W meets only a finite number of transforms γF . Now, let K be a compact subset of D.Choose for each point w of K a fundamental neighborhoodW meeting only a finite numberof transforms γF . Since K is compact, it follows that K meets only a finite number oftransforms γF . Thus (i) holds.

(ii) �⇒ (iii): Take any point w on the boundary of F . It belongs to some boundarysegment B of F . Consider a small fundamental neighborhood W of w. It follows from (ii)

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that W is contained in the union of F and a finite number of transforms γiF containing w.In particular, the part of B in W is also part of the boundary of one of the transforms γiF .Therefore, the part of B in W is part of a side of F . Hence w belongs to a side of F . Thus(iii) holds.

(iii) �⇒ (ii): Consider the union (3.9.1), and its part in a sufficiently small fundamentalneighborhood W of w. The part is the union of a finite number of angular sectors boundedby a finite number (possibly none) of half rays from w. By (iii), we may assume that eachbounding half ray is the part of a side of a transform γiF . However, being part a side of atransform γF , the half ray is also contained in a second transform γ ′F , contradicting the halfray bounded the part inW of the union (3.9.1). Thus there are no bounding half rays. Hencethe union contains W . Thus (ii) holds.

Hence the equivalence of the three conditions has been proved.

(3.10) Definition. The fundamental domainF will be called a normal domain if the equivalentconditions of Proposition (3.9) hold. Note that we have assumed Condition (4) for F . Ingeneral, for an arbitrary fundamental domain, the condition (5i) can be taken as the definitionof normality.

If F is a normal domain, then for the transforms γjF in the union (3.9.1) the sum of theangles of the transforms γjF at w is equal to 2π . Equivalently, the sum of the angles of F atthe points γ−1

j w is equal to 2π . The points γ −1j w are exactly the points v of F ∩ w, and

each v appears as γ−1j w exactly |P w| times. Hence, for a normal domain F , the following

formula holds: ∑v∈F∩ w

AnglevF = 2π/|P w|. (3.10.1)

Conversely, if the equation (3.10.1) holds for all points w, then the condition (5ii) holds, asobserved in (3.7).

(3.11) Proposition. Assume that F is normal fundamental domain for . Then the group P is generated by the boundary transformations γL corresponding to the sides of F .

Proof. Let � be the subgroup of generated by the γL. It suffices to prove the equation,

D :=⋃δ∈�

δF. (3.11.1)

Indeed, assume that the equation holds. Fix a point u of the interior U of F . Let γ be amatrix in . Then γ u belongs to δF for some δ in �. As γ u is an interior point of γF , itfollows that γ , up to ±1, is equal to δ.

To prove the equation (3.11.1), let D0 be the union on the right hand side. Using theproperties of the sides L of F , it follows easily that any point of F is an interior point of afinite union of transforms δF for δ ∈ �. As a consequence, it F meets a transform γF , thenγF = δF for some δ in �.

It follows first that the union D0 is an open subset of D, and next that the complementD − D0 is a union of transforms γD0. Therefore, the complement is open too, and since Dis connected, it follows that D0 = D.

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[Discr] 17

(3.12) Proposition. Let� be a subgroup of finite index d in . Assume for simplicity that�is homogeneous or that is inhomogeneous. Consider a decomposition into right cosets,

= �γ1 ∪ · · · ∪�γd.

Assume that F is a normal fundamental domain for . Then the union of transforms,

G := γ1F ∪ · · · ∪ γdF,

is a normal fundamental domain for �. Moreover, any side of G is of the form γiL whereL is a side of F , and the corresponding boundary transformation is determined as follows:according to the coset decomposition, we have γiγ

−1L = δγk for a unique k and a unique δ

in �. Then δ = δi,L is the boundary transformation corresponding to the side γiL.

Proof. The union of the γiU is an open subset of D, andG is its closure. The interior V ofGcontains the union, and therefore G is the closure of V . The conditions (1)–(3) follow easilyfor G. Moreover, the boundary of G is contained in the union of transforms γiL where L isa side of F . Of course, the inclusion may be strict: some side γiL of γiL may be a side of adifferent γjF and hence not a part of the boundary of G. The assertions of the Propositionfollow easily.

(3.13) Exercise. Modify Proposition (3.12) to cover the case when is homogeneous and� is inhomogeneous.

(3.14) Proposition. Assume that F is a finite fundamental domain. Then F is normal, if andonly if all cusps of F are -parabolic. Moreover, if F is normal, then every -parabolicpoint is -equivalent to a cusp of F and the quotient D/ is compact.

Proof. By hypothesis, there is only a finite number of limit points of F , and they are all cusps.Assume first that they are all -parabolic. To prove thatF is normal, we verify Condition (5i).LetK be a compact subset of D. Then, by Lemma (2.10), each cusp of F has a fundamentalneighborhood U , such that the intersection γU ∩ K is empty for all γ . Clearly, if we cutaway from F a fundamental neighborhood of each cusp, then what remains of F is a compactsubset K1 of F . By Lemma (2.2), the intersection γK1 ∩ K is non-empty for only finitelymany γ . Therefore, the intersection γF ∩ K is non-empty for only finitely many γ . ThusCondition (5i) holds.

Assume conversely that F is normal. Let v be a cusp of F . We have to show that there isin a parabolic matrix with v as fixed point. After a suitable conjugation, we may assumethat (D, v) = (H,∞). Then a fundamental neighborhood of v is a half plane V : z > R.When R is sufficiently big, the intersection F ∩ V is the union of a finite number of verticalstrips, bounded by a finite number of vertical line segments. There is only a finite numberof sides of F . So, enlarging R, we may, by (5iii), assume that each vertical bounding linesegment is part of a side of F . In particular, the rightmost of the vertical bounding segmentsis also part of a side of some transform γ1F different from F . In particular, v = γ1v1 forsome cusp v1 of F . Repeat the argument to the cusp v of γ1F and the rightmost bounding

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vertical segment of γ1F , and continue. As there are only a finite number of cusps of F , itfollows that the given cusp v = ∞ is a fixed point of some matrix γ �= ±1 of .

We claim that γ is parabolic. Indeed, otherwise γ would have a second fixed point differentfrom ∞ on the boundary of H. After a conjugation we may assume that the second fixedpoint of γ is 0. Hence, γ z = rz for some positive real number r �= 1. Replacing γ byγ−1 if necessary, we may assume the r < 1. Now, F contained a vertical strip, bounded bytwo vertical lines and (below) by a horizontal straight line. Clearly, when the powers γ n areapplied to the strip, the two bounding vertical lines move towards the imaginary axis and thebounding straight horizontal line moves towards the real axis. Hence, if K is any compactneighborhood of a point on the imaginary axis, then the intersection K ∩ γ nF is non-emptyfor n � 0. The latter property contradicts Condition (5i). Therefore, γ is parabolic. Thuswe have proved conversely that if F is normal, then every cusp of F is -parabolic.

Assume that F is normal. Denote by F ∗ the union of F and the limit points of F . Thereis only a finite number of limit points. Therefore F ∗ is a compact subset of D ∪ ∂D. Eachcusp is -parabolic, and hence F ∗ is contained in D∪ ∂ D. Consider the image of F ∗ in thequotient D/ . As F ∗ is compact, the image is compact. Moreover, as F contains a systemof representatives for the -orbits in D, the image contains the open subset D/ . The lattersubset is dense in D/ by (2.13). Hence the image of F ∗ is all of D/ . It follows first thatevery point of D ∪ ∂ D is -equivalent to a point of F ∗, and next that the quotient D/ iscompact. Thus the two remaining assertions of the Proposition have been proved.

(3.15) Definition. A discrete subgroup of SL(D) such that the quotient D/ is compactis called a Fuchsian group of the first kind. It follows from Corollary (3.14) that if has afinite normal fundamental domain, then is of the first kind. Conversely, we prove in thenext section that if is of the first kind, then there exists a finite normal fundamental domainF for .

(3.16) Definition. For the limit points of F there is no notion corresponding to the angle atthe points of F . However, for a -parabolic cusp v of F we can define the width of F at vas follows: In a fundamental neighborhood V of v, the intersection F ∩ V is a finite unionof cuspidal sectors. Assume first that (D, v) = (H,∞). Then V is a half plane V : z > R

in H, and the cuspidal sectors are vertical strips. The canonical generator at v = ∞ is atranslation z �→ z+h. Define the width at ∞ as 1/h times the sum of the euclidean widths ofthe strips (where the euclidean width of a vertical strip is the euclidean distance between itsbounding vertical lines). It follows easily from Condition (3) that the width is at most equalto 1.

In general, choose a conjugation α : (D, v) → (H,∞), and define the width of F at the -parabolic point v as the width at ∞ of αF with respect to the conjugate subgroup α . Adifferent choice of α differs from the first choice by a Möbius transformation of the formz �→ rz+ b, where r > 0 and b ∈ R. Hence, for the second choice, the widths of the verticalstrips and the number h are both multiplied by r , and consequently, the quotient is unchanged.Hence the width of F at a -parabolic cusp v is well defined.

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[Discr] 19

If F is a finite normal domain, then following formula holds for any -parabolic point w:

∑v∈F ∗∩ w

widthvF = 1, (3.15.1)

where F ∗ is the union of F and the cusps of F . Indeed, consider the cusps vi ∈ F ∗ that are -equivalent tow, and choose for each a matrix γi such that γivi = w. Then every transformγF having w is a cusp is of the form γ nwγiF , where γw is the canonical generator. Eachtransform γiF is, in a small fundamental neighborhood V of w, a union of cuspidal sectorsbounded by vertical line segments. By (5iii), we may assume that each bounding vertical linesegment is part of a side of γ iF . It follows that the transform γ nwγiF cover all of V . Theformula (3.15.1) is an easy consequence.

Note that, using the canonical generator γv at a point v of F , we can normalize the angleat v: the canonical generator γv is a rotation around v by the angle 2π/|P u|. Define thewidth of F at v to be the angle of F at v divided by the rotation angle of γv . Then the formula(3.15.1) hold for all points w of D ∪ ∂ D. In fact, for points w ∈ D, (3.15.1) is obtainedfrom (3.10.1) by division by 2π/|P w|.

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4. Canonical fundamental domains.

(4.1) Setup. Fix a disk D and a discrete subgroup of SL(D). Let b be a point of D. Thecorresponding canonical domain F(b) is the subset of D consisting of all points z satisfyingthe inequalities,

F(b) : dist(z, b) ≤ dist(z, b′) for all b′ ∈ b. (4.1.1)

We will show in this section that if the point b is -ordinary (i.e., not -elliptic), then theset F(b) is a fundamental domain for . In addition, the interior of F(b) is the set U(b) ofpoints z for which the inequalities (4.1.1) are strict for b ′ �= b. Moreover, we prove that F(b)is a convex polygon with (in general) an infinite number of sides. If is Fuchsian group ofthe first kind, then F(b) is a finite normal fundamental domain.

For the remainder of the section we fix a -ordinary orbit B, say B = b0 where b0is not -elliptic. We consider exclusively the canonical domains F(b) for points b ∈ B.Clearly, for any matrix γ in , we have that γF(b) = F(γ b). Thus the domains F(b) arethe transforms under of the domain F := F(b0) for any given point b0 in B.

(4.2) Observation. The inequality (4.1.1), for two different points b, b ′ of B, defines a(non-euclidean) half plane Fb,b′ of D, bounded by the line,

Lb,b′ : dist(z, b) = dist(z, b′),

of points of equal distance to b and b′. Each half plane Fb,b′ is closed and convex. In addition,if v is any point of Fb,b′ then the open line segment from b to v is contained in the interior ofFb,b′ , that is, for points z on the open line segment, the inequality (4.4.1) is strict.

The subset F(b) is the intersection of the closed half planes Fb,b′ for all b′ �= b in B.Therefore, the subset F(b) is closed and convex in D. In addition, ifw belongs to F(b), thenthe open line segment from b to w is contained in U(b). The latter property holds also whenw is a limit point of F(b), since a point w of ∂D is a limit point of F(b) if and only if it is alimit point of each half plane Fb,b′ .

(4.3) Definition. Fix b0 ∈ B and set F := F(b0). Obviously, if b �= b0, then the intersectionF ∩ F(b) is contained in the line Lb0,b. Moreover, the intersection is closed and convex.Hence the intersection is either empty, or a single point, or a line segment of D (possibly withone or both end points on the boundary ∂D). By definition, a line segment L of D, which isan intersection L = F ∩ F(b) for some b �= b0 in B, is called a side of F . An end point of aside is called a vertex. The finite vertices belong to F , the infinite vertices are limit points ofF .

The line Lb0,b is orthogonal to the line segment from b0 to b and intersects the linesegment in its midpoint. Hence, when b0 is fixed, the line Lb0,b determines the point b. Asa consequence, different sides of F lie on different lines. It follows easily that a finite vertexof F belongs to exactly two sides of F and is a common end point of the two. In addition, aninfinite vertex can belong to at most two sides. A limit point of F which is the common endpoint of two different sides is called a cusp of F .

We prove below that the boundary of F is the union of the sides of F .

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(4.4) Observation. Consider a point u in D. By Lemma (2.11), any fundamental neighbor-hood of u contains only a finite number of points from B. It follows that among the pointsin B there is a finite number of points bj for which the distance to u is minimal. They aresaid to be the points of B nearest to u. They lie on geodesic circle around u (including thepossibility of the circle with radius zero, when u belongs to B). If u ∈ B, there is only asingle b1 := u nearest to u. If u /∈ B, then, according to an orientation of the geodesic circle,we can index the bj in a cyclic order b1, . . . , bn, so that there are no bj on the open arc frombi to bi+1. Then, clearly, any fundamental neighborhood V of u is divided into finitely manyangular sectors,

Vi := {z ∈ V | dist(z, bi) ≤ dist(z, bj ) for all j }, (4.4.1)

and two different Vi and Vj have a half ray from u in common when j = i ± 1, and only thepoint u in common otherwise. If there is only one point bi , then V1 = V , and the angle of V1at u is equal to 2π . If there are two points bi , then V1 and V2 have a line segment as commonboundary. The angles of V1 and V2 at u are equal to π . Finally, when there are more thantwo points bi , then each Vi has at u an angle strictly less than π . In any case, the sum of theangles of Vi at u is equal to 2π . It is convenient to define the width of Vi as the angle of Vidivided by the rotation angle of the canonical generator γu.

Consider next a point u in ∂ D. Then, again by Lemma (2.11), there is a smallest horocycle around u containing points of B. The points of B on this smallest horo cycle are saidto be nearest to u. The set of points of B nearest to u is a discrete subset of the horo cycle.Moreover, it is an infinite set, since it is invariant under the canonical generator γu. Hencethe set of points ofB nearest to u can be indexed cyclically bi for i ∈ Z so that there are no bjon the arc from bi to bi+1. Then, clearly, any fundamental neighborhood V of u is divided ininfinitely many cuspidal sectors Vi by the equations (4.4.1), and two different Vi and Vj havea half ray from u in common when j = i ± 1, and only the point u in common otherwise.

Define the width at u of the cuspidal sector Vi as follows: Assume first that (D, u) =(H,∞). Then the canonical generator γu is a translation z �→ z + h, and the points bi of Bnearest to ∞ are on a horizontal straight line of H. The fundamental neighborhood V is ahalf plane z > R in H, and the cuspidal sector Vi is a vertical strip of H, bounded by thetwo vertical lines of equal distance to bi−1 and bi and to bi and bi+1. Define the width of Viat u = ∞ as the euclidean width of the strip divided by h. In general, define the width of acuspidal sector Vi by using a conjugation as in (3.16).

Note that the width of a cuspidal sector Vi is at most equal to 1, since the set of nearestpoints is stable under the canonical generator γu.

(4.5) Lemma. Let u be a point of D∪ ∂ D, and γu the canonical generator of u. Consider,in the setup of (4.4), a fundamental neighborhood V of u and the division of V in sectors Vi .Then there is a smallest positive number d such that γuVi = Vi+d for all i. Moreover, thesum of the widths of Vj for j = 1, . . . , d is equal to 1.

Proof. If u ∈ D, then any geodesic circle around u is invariant under γu. In particular, γupermutes the points bi . If u ∈ ∂ D, then any horo cycle around u is invariant under γu. In

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[Discr] 23

particular, γu permutes the points bi . It follows, in both cases, that γu permutes the sectorsVi .

Assume that u ∈ D. Then γu is a rotation by the angle 2π/|P u| around u. So thenumber d ≤ n defined by γuVi = Vi+d is independent of i. Clearly the union of the Vj , forj = 1, . . . , e is itself an angular sector, and its angle at u is the rotation angle of γu. Theassertions of the Lemma follow easily.

Assume that u ∈ ∂ D. After a conjugation, we may assume that (D, u) = (H,∞). Thenγu is a translation z �→ z + h, and the cuspidal sectors are vertical strips. So, the number ddefined by γuVi = Vi+d is independent of i. Clearly the union of the Vj , for j = 1, . . . , d,is itself a vertical strip of euclidean width h. The assertions of the Lemma follow easily.

(4.6) Proposition. Let u be a point ofD∪∂ D. Consider, in the setup of (4.4), a fundamentalneighborhood V of u and the division of V in sectors Vi . If V is sufficiently small, thenF(bi) ∩ V = Vi and if b is a point of B such that F(b) ∩ V �= ∅, then b is one of the bi .

Proof. The points bi are the points of B nearest to u. To prove the Proposition, it sufficesto show that a sufficiently small fundamental neighborhood V of u has the following specialproperty: for any point w in V , the points of B nearest to w is a subset of the bi . Indeed,assume that V has the special property. Consider a given i, say i = 0. Obviously, theinequalities in (4.1.1) for b := b0 and all b′ ∈ B imply the inequalities in (4.4.1) for all j .Hence, F(b0) ∩ V ⊆ V0. Conversely, assume that w belongs to V0. By the special property,the points of B nearest to w are among the bi . By definition of V0, among the bi , thenpoint b0 is nearest to w. Hence b0 is a point of B nearest to w, that is, w ∈ F(b0). ThusF(b0)∩ V = V0. Moreover, if the intersection F(b)∩ V is non-empty, take a point w in theintersection. Then b is a point of B nearest to w. So, by the special property, b is one of thebi .

To prove the existence of a fundamental neighborhoodV with the special property, assumefirst that u belongs to D. Let r be the common (non-euclidean) distance from u to the nearestpoints bi . By (2.11), any geodesic disk around u contains only a finite number of points ofB. Hence, if ε > 0 is sufficiently small, then the geodesic diskW with radius r + ε containsof points of B only the bi . The geodesic disk V around u with radius ε/2 has the specialproperty. Indeed, let w be a point of V . If w belongs to V , then, by the triangle inequality,the distance from w to any bj is strictly less than r + ε/2, and the distance from w to a pointoutsideW is at least equal to r + ε/2. Therefore, the points of B nearest to w are among thebi . Thus V has the special property.

Assume next that u is a -parabolic point. After a conjugation, we may assume that(D, u) = (H,∞). Then the canonical generator γu is a translation z �→ z + h, and thesmallest horo cycle containing the nearest points bi is a horizontal (straight) line z = r .It follows from (2.11) that for some r ′ < r , the fundamental neighborhood W : z > r ′contains of points of B only the bi . Let ε be the non-euclidean distance between the straightlines z = r and z = r ′. It is the distance between any two points with the same realpart on the two lines; in fact, it is equal to log(r/r ′). The distance dist(z, z + h) convergesto zero uniformly as z → ∞. Therefore, there exists an R > r such that if z > R and0 ≤ |k| ≤ h/2, then dist(z, z + k) < ε. The fundamental neighborhood V : z > R has

47


the special property. Indeed, let w be a point of V . Let s be the non-euclidean distance fromw to the line z = r (then s = log(u/r)). Since the set of points bi is invariant under thetranslation z �→ z+ h, there is a bj such that for the difference of real parts, k := bj − u,we have that 0 ≤ |k| ≤ h/2. Now, the distance from w to w+ k is strictly less than ε and thedistance from w + k to bj is equal to s. Hence the distance from w to bj is strictly less thans+ ε. On the other hand, the distance from w to the line z = r ′ is equal to s+ ε. Hence thedistance from w to any point of B different from the bi is at least equal to s + ε. Therefore,the points of B nearest to w are among the bj . Thus V has the special property.

(4.7) Observation. Fix a point b0 ∈ B, and set F := F(b0). The local descriptions in (4.5)and (4.6) apply in particular to points v that are either in F or -parabolic limit points of F .It follows that a sufficiently small fundamental neighborhood V of v decomposes into a finitenumber of sectors Vi , of which one, say V0, is equal to F ∩ V . The width of V0 at v is calledthe width of F at v. If v is in F , then the sectors are angular and there is finite number. If vis -parabolic, then the sectors are cuspidal and there is an infinite number.

Consider the following (exhaustive) cases:(1) There is only one Vi . Then V = V0 is contained in F , and hence v is an inner point of

F . Moreover, if F(b) meets V , the b = b0. In particular, no point of V is a side of F or avertex of F . The width of F at v is equal to 1.

(2) There are two angular sectorsVi , sayV0 andV1. ThenV1 = F(b1)∩V , andV ∩F(b) =∅ when b is not b0 or b1. The common boundary of V0 and V1 is the intersection of V andthe line Cb0,b1 , and equal to the part in V of the intersection F(b0) ∩ F(b1). In particular,the latter intersection is a side of F , it contains v, and it is the only side of F containing v.The width of F at v is equal to 1/2. The interior points of V0 are interior points of F . Inparticular, the point v belongs to the closure of the set of interior points of F .

(3) There are more than two angular sectors Vi . Then the common boundaries of V0 andthe two adjacent Vi’s are parts of sides of F . In particular, then v is a vertex of F . Thewidth of F at v is strictly less than 1/2. The interior points of V0 are interior points of F . Inparticular, the point v belongs to the closure of the set of interior points of F .

(∞) Assume that v is a -parabolic limit point of F . If V−1 and V1 are the two adjacentVi , then the intersections V0 ∩ V−1 and V0 ∩ V1 are parts of two sides of F . Thus v is thecommon end point of two different sides; in particular, then v is a cusp of F .

(4.8) Theorem. Let b0 be a point of D which is not -elliptic. Then the canonical domainF = F(b0) of (4.1) is a fundamental domain for , and its interior is the subset U(b0). Theboundary of F is the union of the sides of F . In addition, every -parabolic limit point of Fis a cusp. Finally, every compact subset of D meets only a finite number of sides.

Denote by F ∗ the union of F and the set of -parabolic limit points of F . Then, for anypoint u of D ∪ ∂ D, the intersection F ∗ ∩ u is finite, and the following formula holds,

∑v∈F ∗∩ u

widthvF = 1

In particular, every -parabolic point is equivalent to a point in F ∗.

48


[Discr] 25

Proof. Set U := U(b0), and consider the conditions (1)–(3) of (3.1). It follows from Propo-sition (4.6) in particular that any point v of D belongs F(b) for some b = γ b0, and that itbelongs to only a finite number of F(b). As F(γ b0) = γF(b0), it follows that condition(1) holds. Next, let v be a point in U . Then b0 is the single point in B = b0 nearest tov. It follows from the discussion in (4.7)(1) that a sufficiently small neighborhood of v iscontained in F . Hence U is an open subset of D. If a point of F is not in U , it follows fromthe discussion that any neighborhood of v contains points ofU and points of the complementof F . Hence U is the set of interior points in F and F is the closure of U . Thus condition (2)holds. Finally, the intersection U ∩ γU is equal to U(b0) ∩ U(γ b0). It is obvious from thedefinition in (4.1) that the intersection is empty unless b0 = γ b0. Since b0 is not -elliptic,the equation b0 = γ b0 implies that γ = ±1. Hence condition (3) holds. Thus F is afundamental domain for .

The assertions about the boundary of F follow immediately from the discussion in (4.7).By the same discussion, any point u of D has a fundamental neighborhood meeting at mosttwo sides of F . It follows that a compact subset of D can only meet a finite number of sidesof F .

Finally, the formula follows from Lemma (4.5).

(4.9) Corollary. The quotient D/ is compact, if and only if F = F(b0) is a finite normaldomain.

Proof. By Proposition (3.14), if the fundamental domain F is a finite normal domain, thenthe quotient is compact.

Conversely, assume that the quotient is compact. Choose for each point u of D ∪ ∂ Da small fundamental neighborhood V = Vu of u having the property of Proposition (4.6).If V has the property for u, and u′ is -equivalent to u, say u′ = γ u, then V ′ := γV hasthe property for u′. Moreover, V ′ is independent of the choice of γ , because two differentchoices differ by a matrix in u, and the matrices of u leaves the fundamental neighborhoodV invariant. Hence we may assume that Vγu = γVu for all γ in .

For any point v of F ∗, the fundamental neighborhood Vv is a union of sectors Vi . Let Fvbe the sector contained in F ∗.

Consider an arbitrary point u of D ∪ ∂ D and the fundamental neighborhood Vu. ThenVu is a union of sectors Vi and Vi is the part in Vu of F(bi). We have that bi = γ b0 for someγ in , and hence F(bi) = γF . Thus, with v := γ −1u, we have that v ∈ F ∗ and Vu = γVv.It follows that Vi is equal to γFv . Therefore, in the decomposition of Vu, any of the sectorsVi is a transform of Fv for some point v ∈ F ∗ ∩ u.

The intersection F ∗ ∩ u is finite by the Theorem. Hence it follows from the precedingargumentation that any point u in D∪∂ D has a fundamental neighborhood which is a unionof transforms of sets Fv for a finite number of points v in F ∗. It follows that any point inthe quotient has a neighborhood which is a finite union of the images of the Fv . Since thequotient is compact, therefore the quotient is the union of a finite number of images of the Fv .In other words, there is a finite set of points vk in F ∗ such that any point in F ∗ is -equivalentto a point in the union of the Fvk .

49


Let w be a point which is either on the boundary of F or a limit point of F . LetM denotethe open line segment from b0 to w. Then M is contained in the interior of F . In particular,no points of M can belong to a transform of F different from F . Hence M is contained inthe union of the finitely many Fvi . It follows first that w is a boundary point of one of theFvk , and next that w belongs to one of the Fvk .

Each Fvk meets at most two sides of F and has at most one point on ∂D. As the boundaryof F is the union of the sides, it follows first that the number of sides is finite. Hence thecondition (4) of (3.1) holds. Next, it follows that there is only a finite number of limit pointsfor F . Hence F is a finite domain. Moreover, the limit points of F are in F ∗ and hence -parabolic. Therefore, by Proposition (3.14), F is a finite normal domain.

Thus the Corollary has been proved.

(4.10) Note. If the canonical domain F = F(b0) has finite area, then it is a finite normaldomain. Indeed, assume that the area of F is finite. For each finite vertex v of F , let αvdenote the angle of F at v (equal to 2π times the width at v). Set αv := 0 for all limit pointsv of F . Let vi be a finite number of points that are either vertices or limit points of F . Let F ′be the convex hull of the vi . Then F ′ is a finite polygon, possibly with some of its vertices on∂D. If α′

i is the angle of F ′ at vi , then it is well known that the area of F ′ plus 2π is equal to∑(π − α′

i). Now, since F is convex, we have that F ′ ⊆ F . So, the area of F ′ is at most thearea of F and the angle α′

i is at most αvi . Therefore, the sum∑(π − αvi ) is at most equal to

the area of F plus 2π . As a consequence, the following sum over all points v that are eithera vertex or a limit point of F is bounded above:

∑(π − αv). (*)

In the sum (*), each limit point of F contributes with the term π . Hence there is only a finitenumber of limit points of F . Divide the finite vertices into -equivalence classes. Each classis finite, and sum of the angles in a given class is equal to 2π/d where d is the commonnumber of elements of the isotropy group P v for any vertex v in the class. Let n be thenumber of vertices in the class. Then the sum of the terms in (*) corresponding to the verticesin the class is equal to (n − 2/d)π . Each angle is strictly less than π . Therefore, if d = 1,then n ≥ 3, and if d = 2 then n ≥ 2. Hence, n− 2/d ≥ 1 if d = 1 or d = 2. If d > 2, then(n − 2/d) ≥ 1/3. Hence, in all cases, the group contributes with at least π/3 to the sum. Itfollows that there is only a finite number of groups. Hence the number of finite vertices isfinite.

As the number of vertices is finite, so is the number of sides. Moreover, there is only afinite number of limit points of F . So F is a finite domain. Now it follows from Theorem(3.7) that F is a finite normal domain.

(4.11) Exercise. Prove that the considerations in (4.10) imply the following estimate:

Area(F )/π + 2 ≥ (number of infinite vertices)+ (number of finite vertices)/3.

50


[Discr] 27

5. The Euler characteristic.

(5.1) Setup. Fix a disk D and a discrete subgroup of SL(D) such that the quotient X =X( ) := D/ is compact.

The quotient is a surface by Corollary (2.13): any point in the quotient has an openneighborhood topologically isomorphic to an open disk in the plane. Therefore, as is wellknown, the Euler characteristic χ(X) is defined. It is the alternating sum of the ranks of thecohomology groups,

χ(X) = dimH 0(X)− dimH 1(X)+ dimH 2(X). (5.1.1)

Obviously, the quotientX is connected. In addition, the definition of the local isomorphismsin Corollary (2.13) shows thatX in a natural way is an oriented surface. Therefore, as is wellknown, the surface X is (topologically) a sphere with a certain number of handles attached.The number of handles is called the genus ofX and is denoted g = g(X) = g( ). In addition,the cohomology groups H 0(X) and H 2(X) are 1-dimensional, and H 1(X) is of dimension2g. Hence the genus and the characteristic are related by the formula,

χ = 2 − 2g. (5.1.2)

Moreover, it is well known that the characteristic can be determined from any a triangulationof X by the formula,

χ(X) = #(vertices)− #(edges)+ #(faces). (5.1.3)

Fix a finite normal fundamental domainF for , cf. Proposition (3.14) and Corollary (4.9).The boundary of F is the union of finitely many sides. We will take as definition of a vertexhere a point which is an end point of a side. Recall that for any side L of F there is a uniqueboundary transformation γL in such that γLL is a side of F . We will say thatL is an ellipticside if γLL = L. Equivalently, a side L is elliptic, if it contains an elliptic point v which isnot a vertex. Then, necessarily, the point v is elliptic of order 2 and γL is the rotation by theangle π around v.

(5.2) Definition. The Euler characteristic of the domain F is defined from any triangulationof F as the number,

χ(F ) = #(inner vertices)− #(inner edges)+ #(faces). (5.2.1)

The Euler characteristic is in fact a topological invariant of the interior U of F . If U isconnected, then the characteristic is equal to 1 minus the number of holes inU . At any rate, itis a consequence of (the proof of) the next formula that the characteristic of F is independentof the choice of triangulation.

51


(5.3) Proposition.

Area(F ) = −2πχ(F )+ #(sides of F)π −∑

AnglevF, (5.3.1)

where the sum is over the vertices of F and the angle of F at a cusp is defined as 0.

Proof. Consider a triangulation of F . Each face is a triangle. An edge of a triangle is eitheran inner edge or part of the boundary of F . Clearly, if we divide a side of F into two by apoint on the side, then the number of sides is increased by 1 and the sum of the angles of isincreased by π ; hence the right side of (5.3.1) is unchanged. Therefore, we may assume thatthe sides of F are exactly those edges of the triangulation that are contained in the boundary.

If we sum the areas of the faces of the triangulation, we get the area of F , that is the leftside of (5.3.1). On the other hand, each face is a triangle and its area is equal to π minus thesum of the three angle of the triangle. Hence the sum of the areas of the faces is equal to πtimes the number of faces minus the sum of the angles of the faces. In the latter sum, eachinner vertex contributes with 2π and each vertex on the boundary of F contributes with theangle of F at the vertex. Hence we have obtained the equation,

Area(F ) = #(faces)π − #(inner vertices) 2π −∑

AnglevF.

The equation (5.3.1) is a consequence, since obviously, the equation,

3#(faces) = 2#(inner edges)+ #(sides),

holds for a triangulation whose outer edges are the sides of F .

(5.4) Proposition. Take as vertices of F the end points of the sides. Then the Euler charac-teristics of X = D/ and the fundamental domain F are related by the formula,

χ(X) = χ(F )− #(non-elliptic sides of F)

2+ #

(vertices of F

), (5.4.1)

where the last fraction denotes the set of -equivalence classes of vertices of F .

Proof. Consider an elliptic sideL of F . It does not contribute to the right hand side of (5.4.1).The boundary transformation γL is a rotation by π around an elliptic point v of order 2 on L.The point v divides L into two line segments L′ and L′′ with v as common end point. Thetwo segments L′ and L′′ are interchanged by γL. Hence, considering the sides of F , we mayreplace the side L by the two sides L′ and L′′ and add the point v as a vertex. The point vis -equivalent to no other point of F . Therefore, the replacement does not change the righthand side of (5.4.1). Thus, to prove (5.4.1), we may assume that no side of F is elliptic.

Consider a side L and its transform L′ = γLL. A given point v interior on L dividesL into two line segments L1 and L2 with v as common end point. Then L′ is divided intothe two line segments L′

1 := γLL1 and L′2 = γLL2 with v′ := γLv as common end point.

Hence, considering the sides of F , we may replace the two sides L and L′ by the four sides

52


[Discr] 29

L1, L′1, L2, L′

2 and add the two points v and v ′ as a vertices. The points v and v ′ form a single -equivalence class of vertices. Therefore, the replacement does not change the right handside of (5.4.1). Thus, to prove (5.4.1), we can divide any side L of F into a finite numberof sides as long as we divide similarly the side γL. In particular, just dividing some of thesides into two sides if necessary, we may assume that no two different points on a side are -equivalent.

Consider a triangulation T of F . As noted above, we may assume that any edge of Twhich is contained in the boundary of F is a side of F . Then every triangle of T is mappedhomeomorphically onto its image in the quotient X, and the images form a triangulation ofX. Clearly, in the image triangulation, the faces correspond to the faces of T , the edgescorrespond to inner edges of T and pairs L, γLL of sides of F , and the vertices correspondto inner vertices of T and -equivalence classes of vertices of F . Thus the required equation(5.4.1) follows from (5.1.3) and the definition of χ(F ).

(5.5) Corollary. The following formula holds:

Area(F )

2π= −χ(X)+

∑w mod

(1 − 1

|P w|), (5.5.1)

where the sum is over all orbits that are -elliptic or -parabolic, and |P w| is the commonorder of the isotropy groups of the points in an orbit. In the sum, at the -parabolic orbitswhere the isotropy group is infinite, the fraction 1/|P w| is counted as 0.

Proof. We apply the two formulas (5.2.1) and (5.4.1). It is clear, and noted in the proof of(5.4), that to apply (5.4.1) we may assume that no side of F is -elliptic. Hence we obtainthe equation,

Area(F )

2π= −χ(X)+ #

(vertices of F

)−

∑AnglevF /2π.

In the sum, group the terms according to their -equivalence class. Consider a -equivalenceclass of vertices represented by a pointw, that is, a class consisting of all vertices -equivalentto w. If w is in D, then, by (3.10), the corresponding contribution to the sum is equal to1/|P w|. If w is in ∂ D, corresponding to an equivalence class of cusps of F , then allthe angles are zero, and the corresponding contribution in the sum is zero, and hence byconvention equal to 1/|P w|. Hence, we obtain the equation,

Area(F )

2π= −χ(X)+

∑w mod

(1 − 1

|P w|),

where the sum is over all orbits containing a vertex of F . The asserted equation (5.5.1) is aconsequence since, by assumption, every elliptic point is -equivalent to a vertex.

53


(5.6) Note. It follows from (5.5) that the area of the fundamental domain F is always arational multiple of 2π , that is, the quotient Area(F )/2π is a rational number. The quotientis denotedμ = μ( ). In addition, it is customary to denote by νe( ) the number of -ellipticorbits of order e. In addition, the number of -parabolic orbits is denoted ν∞( ). Hence,using (5.1.2), the following equation is just a rewriting of (5.5.1),

μ = 2g − 2 + ν∞ +∑e≥2

νe

(1 − 1

e

). (5.6.1)

(5.7) Observation. If � is a subgroup of finite index in , then

μ(�) = |P :P�|μ( ). (5.7.1)

Indeed, if the transformations γi , for i = 1, . . . , d, represent the right cosets of P moduloP�, then, as observed in (3.12), the union G := ⋃

γiF is a normal fundamental domain for�. Obviously, the area of G is d times the area of F . As a consequence, (5.7.1) holds.

(5.8) Example. The group (1) := SL2(Z) is a discrete subgroup of SL(H). A fundamentaldomain for (1) was determined in (1.6). The domain was a triangle with one cusp and twofinite -equivalent vertices. Obviously for a triangle, and more generally for any polygonwith a simply connected interior, the Euler characteristic is equal to 1. The side connectingthe two finite vertices is elliptic, and there are two equivalence classes of vertices. Hence, bythe formula (5.4.1), the Euler characteristic of (1) is 1− 1+ 2 = 2. Thus the genus is equalto 0, confirming that the quotient H/ (1) obviously is a sphere.

The angles of the fundamental domain at the two finite vertices are equal to 2π/6. Hencethe area of the domain is equal to π/3. In other words, μ = 1

6 . There is one parabolicorbit, one elliptic orbit of order 2 and one of order 3 and no other elliptic orbits. Henceν∞ = ν2 = ν3 = 1, and νe = 0 for e > 3.

Consider a subgroup of finite index in (1). Set d := |P (1):P |. Then μ( ) = d/6by (5.7.1). Moreover, since every -elliptic point is (1)-elliptic, we have νe = 0 for e > 3.Hence, with νe = νe( ), equation (5.6.1) is equivalent to the following,

g( ) = 1 + d

12− ν2

4− ν3

3− ν∞

2. (5.8.1)

(5.9) Note. The area of the fundamental domain F can be obtained by integration. Assumefirst the disk is the upper half plane H. Then,

Area(F ) =∫∂F

dz

z =∫∂F

dx

y=

∫F

dx dy

y2 . (5.9.1)

The two middle path integrals are over the boundary of F , orientated counter clockwisearound F . The first path integral in (5.9.1) should be taken with care. If a side L of F has aninfinite vertex as end point, then the integral

∫Ldz/y is not convergent. When F has infinite

54


[Discr] 31

vertices, the path integral along ∂F is defined as follows: consider the path integrals obtainedby integration along the boundaries of the subdomains obtained from F by cutting away asmall fundamental neighborhood of each infinite vertex. The limit of these path integrals, asthe fundamental neighborhoods shrink around the infinite vertices, is then the path integralalong ∂F .

In the first integral of (5.9.1), the integrand is a sum of two forms,

dz

y= dx

y+ i

dy

y.

The second form is a total differential in H, y−1dy = d(log y). Hence its integral over aclosed path is equal to zero. Therefore, the second equality of (5.9.1) holds.

Consider the integral∫Ldx/y over any oriented line segmentL inH. SayL is the segment

from u to v, where u and v are allowed to be limit points. Then the path integral∫Ldx/y is

convergent, and given by the formula,

∫L

dx

y= θL,v − θL,u, (5.9.2)

where, for any point w on L, the angle θL,w is the oriented angle from L to the (vertical) linefromw to ∞. The formula follows easily from the definition of the path integral, noting thatLis part of a circle orthogonal to the real axis (the trivial case whenL is part of a vertical straightline has to be treated separately). From the formula (5.9.2) it follows that if F is a triangle,then the path integral

∫∂Fdx/y is equal to π minus the sum of the interior angles. Thus the

equation∫∂Fdx/y = Area(F ) holds when F is a triangle. Therefore, using a triangulation

of F , the equation holds in general. As d(y−1dx) = −y−2 dy∧dx = y−2 dx∧dy, the lastequality in (5.9.1) follows from Stoke’s theorem. Thus the equations of (5.9.1) have beenproved.

The first equation of (5.9.1) implies the following:

Area(F ) =∑

i

∫L

J (γL, z)′

J (γL, z)dz. (5.9.3)

where the sum is over the sidesL of F , and γL is the boundary transformation. Indeed, if γ ∈SL(H) then J (γ, z) = cz+ d with real numbers c and d. It follows that 2iJ (γ, z)′ = 2ic =(J (γ, z) − J (γ, z))/z. Moreover, (γ z) = |J (γ, z)|−2z and d(γ z) = J (γ, z)−2dz.Hence we obtain the equations,

2i∫L

J (γ, z)′dzJ (γ, z)

=∫L

(1 − J (γ, z)

J (γ, z)

)dzz =

∫L

( dzz − d(γ z)

(γ z))

=∫L

dz

z −∫γL

dz

z .

Take γ := γL. Then γLL is a side L′ of F . However, the orientation on L′ as a side of Fis the reverse of the orientation on L′ as an image γLL. Hence the difference on the right

55


hand side of the equations is the sum∫Ldz/z + ∫

L′ dz/z. Consequently, (5.9.3) followsby summation over all the sides L of F . In fact, it follows that

Area(F )

2=

∑′i

∫L

J (γL, z)′

J (γL, z)dz. (5.9.4)

where the sum is over unordered pairs {L,L′} of sides with L′ = γLL.It is not hard to see, using a Möbius transformation to transform the integrals, that (5.9.3)

and (5.9.4) holds for any disk D.

56


[Mdlar] 1

Modular groups

1. Finite projective lines.

(1.1) Setup. Recall that the group SL(H) is the group SL2(R). The subgroup SL2(Z) is adiscrete subgroup. It is called the modular group.

Fix a natural number N . We are going to consider various subgroups of SL2(Z) relatedto subgroups of SL2(Z/N). Note that, by the Chinese Remainder Theorem, if N = ∏

pν isthe prime factorization of N , then

GL2(Z/N) =∏

GL2(Z/pν), SL2(Z/N) =∏

SL2(Z/pν).

Clearly, the definition of the action of GL2(C) on the Riemann sphere C extends to anarbitrary field. More generally, if R is any commutative ring, we define PGLn(R) as thequotient of GLn(R) modulo the scalar matrices:

PGLn(R) := GLn(R)/R∗.

By definition, PSLn(R) is the image of SLn(R) in PGLn(R), that is,

PSLn(R) := SLn(R)/μn(R),

where μn(R) is the subgroup of R∗ consisting of n’th roots of unity. Note in particular, forn = 2, that the group PSL2(R) is in general a non-trivial quotient of SL2(R)/± 1, since theequation α2 = 1 in R may have non-trivial solutions.

Assume that R is a local ring with maximal ideal m. Let R be the disjoint union, R :=R ∪ 1

m, where 1

mis a set consisting of one symbol for each element in m. Denote by R2 the

R-module of columns, and by (R2)∗ the subset of columns for which at least one entry is aunit. Then there is a surjective map (R2)∗ → R,[

z1z2

]�→ z1/z2. (1.1.1)

The quotient z1/z2 ∈ R is defined as z1z−12 if z2 ∈ R∗ and as the symbol 1/(z−1

1 z2) ifz2 ∈ m. Obviously, the group GL2(R) acts on the set (R2)∗ and the scalar matrices permutethe elements of the fibers of the map (1.1.1). Hence the action descends to an action ofPGL2(R) on R. As in the case of Möbius transformations, it is a faithful representation,

PGL2(R) ↪→ Aut(R). (1.1.2)

The set R for a local ring R is the projective line IP 1(R). The action of PGL2(R) on theprojective line generalizes to an action of PGLn+1(R) on projective n-space IP n(R) :=(Rn+1)∗/R∗.

57

[Mdlar] 2 Automorphic functions26. februar 1995

(1.2) Note. Consider a field F . Then it follows, as in the proof of (Möb.1.3)(3), that theaction of PGL2(F ) on F is triply transitive. However, the action of the subgroup PSL2(F )

is in general only doubly transitive: if (u, v) and (u′, v′) are two pairs of different points ofF then the matrices α that map the first pair to the second are the matrices of the followingform (with an obvious notion for the representatives of the four points):

α = [u′ v′][λ 00 μ

][u v]−1,

where λ, μ ∈ F ∗. Clearly, with suitable choices of λ and μ, the determinant of α is equal to1. For instance, if F is a finite field with q elements, then there are (q − 1)(q − 1) matricesα, and q − 1 of these have determinant 1. So, if q is odd, there are (q − 1)/2 automorphismsin PSL2(F ) mapping the first pair to the second.

(1.3) Proposition. Consider in SL2(Z) the three matrices,

s :=[ 0 −1

1 0

], t :=

[ 1 10 1

], u :=

[ −1 −11 0

].

Then s2 = −1, u3 = 1, and s = tu. In particular, s is of order 4 and u is of order 3.Moreover, the group SL2(Z) is generated by the two matrices s and t , and hence also by sand u.

Proof. The three asserted equations follow by a simple computation. In addition,

tb =[ 1 b

0 1

], st−1s−1 =

[ 1 01 1

]. (1.3.1)

Denote by the subgroup generated by s and t . It follows from (1.3.1) that the elementaryrow and column operations on matrices in GL2(Z) can be achieved by multiplication fromthe left and right by matrices in . By the Euclidean algorithm, every matrix in GL2(Z) canbe changed into a diagonal matrix by the elementary operations. In particular, any matrix σin SL2(Z) can, by multiplications from the right and left by matrices in , be changed into±1. As −1 = s2, it follows that σ belongs to .

(1.4) Sublemma. The group SL2(Z/N) is generated by the following matrices:

s :=[ 0 −1

1 0

], t :=

[ 1 10 1

], dμ :=

[μ 00 μ−1

]for μ ∈ (Z/N)∗. (1.4.1)

Proof. Note first that if the integers a, c,N are relatively prime, then there is a number k suchthat a + kc is relatively prime to N . Indeed, it suffices to take k := ∏

p, where the productis over all primes p such that p is a divisor of N and not a divisor of a.

Consider an arbitrary matrix σ in SL2(Z/N):

σ =[a b

c d

],

58


[Mdlar] 3

where ad − bc ≡ 1 (mod N). We have to show that we can obtain the identity matrix bymultiplying σ a finite number of times by the matrices in (1.4.1) and their inverses.

Clearly, the integers a, c, N are relatively prime. Hence there exists a number k such thata+kc is prime toN . Therefore, replacing σ by t kσ , we may assume that a ∈ (Z/N)∗. Next,replacing σ by d−1

a σ , we may assume that a = 1. Furthermore, since

σ t−b =[ 1 b

c d

][ 1 −b0 1

]=

[ 1 0c 1

],

we may assume that a = d = 1 and b = 0. Finally,

sσ s−1t c = s[ 1 0c 1

]s−1t c =

[ 1 −c0 1

]t c = 1,

and hence the identity matrix has been obtained, as required.

(1.5) Proposition. Reduction modulo N is a surjective homomorphism of groups,

SL2(Z) → SL2(Z/N).

As a consequence, the group SL2(Z/N) is generated by the first two matrices s and t of(1.4.1).

Proof. To prove that reduction modulo N is surjective, it suffices to lift the generators of(1.4.1). Obviously s and t lift. For a diagonal matrix dμ, we have that μ = m, where m isprime toN . Thenm is prime toN 2. Hence there are numbers x and y such that xm−yN2 = 1.Clearly, the following matrix of SL2(Z) is a lift of dμ:

[m yN

N x

].

Hence the reduction is surjective. It follows that SL2(Z/N) is generated by s and t , sincethe lifts of s and t generate SL2(Z) by (1.3).

(1.6) Corollary. The group GL2(Z/N) is generated by the matrices,

s =[ 0 −1

1 0

], t =

[ 1 10 1

], dμ,1 :=

[μ 00 1

]for μ ∈ (Z/N)∗. (1.6.1)

Proof. Clearly, the assertion follows from the last assertion of Proposition (1.5).

(1.7) Corollary. Letp be a prime number. Then, forν ≥ 1, the surjective ring homomorphismZ/pν → Z/p induces a surjective group homomorphism,

GL2(Z/pν) → GL2(Z/p), (1.7.1)

59


and the kernel of the induced homomorphism consists of all matrices in Mat2(Z/pν) of theform,

1 + α, (1.7.2)

where all entries in α belong to the kernel of Z/pν → Z/p.

Proof. Clearly, the generators s and t of (1.6.1) for N := p lift, and the generator d1,μ lifts,because (Z/pν)∗ → (Z/p)∗ is surjective. Hence the homomorphism (1.7.1) is surjective.Obviously, any matrix in the kernel has the form (1.7.2). Conversely, for a matrix σ = 1+α ofthe form (1.7.2), the determinant is congruent to 1 modulop; hence σ belongs to GL2(Z/pν),and obviously σ belongs to the kernel of (1.7.1).

(1.8) Proposition. The orders of the groups GL2(Z/N) and SL2(Z/N) are given by thefollowing formulas,

| GL2(Z/N)| = N4 ∏p|N(1 − 1

p)(1 − 1

p2 ), | SL2(Z/N)| = N3 ∏p|N(1 − 1

p2 ).

As a consequence,

| SL2(Z/N)/± 1| ={

6 for N = 2,12N

3 ∏p|N(1 − 1

p2 ) for N > 2.

Proof. By the Chinese Remainder Theorem, it suffices to prove the formulas when N is aprime power,N = pν . The quotient Z/p is the field Fp with p elements. Hence the order ofthe group GL2(Z/p) is equal to the number of bases of the 2-dimensional vector space F 2

p ,

that is, the order is equal to (p2 −1)(p2 −p). By Lagrange, the kernel of the homomorphismZ/pν → Z/p is of order pν−1. Hence, there are (pν−1)4 matrices of the form (1.7.2).Consequently, by Corollary (1.7),

| GL2(Z/pν)| = (pν−1)4(p2 − 1)(p2 − p) = (pν)4(1 − p−2)(1 − p−1).

Thus the first asserted formula holds. The second formula follows from the first, becauseSL2(Z/pν) is the kernel of the surjective homomorphism,

det : GL(Z/pν) → (Z/pν)∗,

and (Z/pν)∗ has order pν − pν−1 = pν(1 − p−1).Clearly, the final formula is a consequence, since 1 = −1 in Z/N only when N = 2 (or

N = 1).

(1.9) Remark. The order of P SL2(Z/N) is obtained from the order of SL2(Z/N) by dividingby the order of the groupμ2(Z/N) defined in (1.1). Clearly, the latter order is a multiplicativefunction of N . Obviously, for an odd prime power pν , the order of μ2(Z/pν) is equal to 2.For a power of 2, the order of μ2(Z/2ν) is equal to 1 for ν = 1, equal to 2 for ν = 2 andequal to 4 for ν ≥ 3.

60


[Mdlar] 5

(1.10) Lemma. (1) The group SL2(Z/1) is the trivial group of order 1.(2) The group PSL2(Z/2) = SL(Z/2) is the dihedral group D3 = S3 of order 6.(3) The group PSL2(Z/3) = SL(Z/3)/± 1 is the tetrahedral group A4 of order 12.(4) The group PSL2(Z/4) = SL(Z/4)/± 1 is the hexahedral group S4 of order 24.(5) The group PSL2(Z/5) = SL(Z/5)/± 1 is the dodecahedral group A5 of order 60.(6) The group PSL2(Z/6) = SL(Z/6)/± 1 is isomorphic to the group S3 × A4 of order

72.

Proof. The assertion (1) is trivial. By the Chinese Remainder Theorem, the assertion (6)follows from (2) and (3). To prove the remaining assertions, consider for N = 2, 3, 4, 5the ring Z/N . Then Z/N is a field for N = 2, 3, 5, and a local ring for N = 4. In eachof the four cases, the order of the group PSL2(Z/N) is given by the formula of (1.8). Toidentify the group with the asserted permutation group, consider the projective line Z/N andthe representation of (1.1.2),

PSL2(Z/N) ↪→ Aut(Z/N). (1.10.1)

N = 2: The group PSL2(Z/2) has order 6. The projective line Z/2 has 3 elements:∞, 0, 1. Hence the right hand side of (1.10.1) is the symmetric group S3. Thus the two sideshave the same order, and hence the inclusion is an isomorphism.N = 3: The group PSL2(Z/3) has order 12. The projective line Z/3 has 4 elements:

∞, 0, 1,−1. Hence the right hand side of (1.10.1) is the symmetric group S4. As the lefthand side is of order 12, it is equal to the unique subgroup A4 of index 2 in S4.N = 4: The group PSL2(Z/4) has order 24. The projective lineX := Z/4 has 6 elements:

∞, 12 , 0, 2, 1,−1, and there is an obvious map Z/4 → Z/2. Given any point x in X there

is a unique second point x ′ having the same image as x in Z/2. Consider subsets Z ofX consisting of 3 elements lying over the 3 elements ∞, 0, 1 of the projective line Z/2.Obviously, if Z = {x, y, z} is such a subset, then so is the complement Z ′ = {x ′, y ′, z′}.Hence, there are 4 elements in the set D of all decompositions of X into two such subsets,

X = {x, y, z} ∪ {x ′, y ′, z′}.

Clearly, the group PSL2(Z/4) acts on the set D. Consequently, we obtain a representation,

PSL2(Z/4) → Aut(D) = S4. (1.10.2)

The representation (1.10.2) is injective. Indeed, assume that σ in SL2(Z/4) acts as the identityon D. Take a decomposition in D, say {a, b, c} ∪ {a ′, b′, c′}. Then {a, b′, c′} ∪ {a′, b, c}is a second decomposition. As the first decomposition is fixed under σ , we have eitherσ {a, b, c} = {a, b, c} orσ {a, b, c} = {a′, b′, c′}. Assume thatσ {a, b, c} = {a′, b′, c′}. Then,since the second decomposition is fixed under σ , it follows first that σ {a, b ′, c′} = {a′, b, c}and next that σa = a′. Similarly, σb = b′ and σc = c′, and hence σx = x′ for allx. However, it is easily verified that the permutation x �→ x ′ can not be obtained from

61


a matrix with determinant 1. So the case σ {a, b, c} = {a ′, b′, c′} is excluded. Therefore,σ {a, b, c} = {a, b, c}. As the second decomposition is fixed under σ , it follows that σa = a

and σa′ = a′. Similar equations are obtained for b and c, and hence σ is the identity inPSL2(Z/4).

As the representation (1.10.2) is injective and the two groups have the same number ofelements, the representation is an isomorphism.N = 5: The group G = PSL2(Z/5) has order 60. The projective line X := Z/5 has

6 elements: ∞, 0, 1, 2,−2,−1. The action of G on X is doubly transitively. In fact, asobserved at the end of (1.2), if (u, v) and (u′, v′) are any two pairs of different points ofX, then there are two elements of G mapping the first pair to the second. Take a subset{x1, x2} with two elements. There are two elements ofG for which x1 and x2 are fixed points.One element is the identity, and the second is (as seen by reducing to the case x1 = ∞and x2 = 0) a double transposition (y1, y2)(z1, z2). Similarly, there are two elements ofG that interchanges x1 and x2, namely the two double transpositions (x1, x2)(y1, y2) and(x1, x2)(z1, z2). Therefore, the subset {x1, x2} is part of a unique decomposition of X intothree parts,

X = {x1, x2} ∪ {y1, y2} ∪ {z1, z2}, (1)

with the special property that any element in G that stabilizes one part will stabilize all threeparts. There are 15 subsets with two elements. Hence, there are 5 elements in the set Dof all decompositions (1) with the special property. Clearly, the group G acts on the set D.Consequently, we obtain a representation,

PSL2(Z/5) = G → Aut(D) = S5. (1.10.3)

The representation is injective. Indeed, assume that α is a nontrivial element ofG that acts asthe identity on D. Take a point x1 such that αx1 �= x1. Consider first the decomposition (1)with x2 := αx1. Since the decomposition is invariant, it follows that α{x1, x2} = {x1, x2}.Hence α has two fixed points, namely either y1 and y2, or z1 and z2. Consider next thedecomposition (1) with a fixed point of α as x2. Then, again since the decomposition isinvariant, it follows that α{x1, x2} = {x1, x2}, which is a contradiction since αx1 �= x1 andαx2 = x2. Hence the representation is injective.

As the representation (1.10.3) is injective, necessarily its image is the unique subgroupA5of index 2 in S5.

Hence all the isomorphisms of the Lemma have be established.

(1.11) Exercise. (1) The projective line X := Z/3 has 4 elements: ∞, 0, 1, −1. Provethat, as permutations of X, we have s = (∞, 0)(1,−1) and t = (0, 1,−1). The tetrahedralgroup T is the automorphism group of a tetrahedron. The elements of T permute the 4 facesof the tetrahedron, and so T is a subgroup of S4. Prove that the permutations s and t are‘tetrahedral’, that is, under a suitable labeling of the four faces, s and t can be realized asrotations of the tetrahedron. Conclude that PSL2(Z/3) = T .

(2) The projective line X := Z/4 has 6 elements: ∞, 12 , 0, 2, 1, −1. Prove that, as

permutations of X, we have s = (∞, 0)(2, 12 )(1,−1) and t = (0, 1, 2,−1). The hexahedral

62


[Mdlar] 7

groupH is the automorphism group of a hexahedron. The elements ofH permute the 6 facesof the hexahedron, and so H is a subgroup of S6. Prove that the permutations s and t are‘hexahedral’, and conclude that PSL2(Z/4) = H .

(3) The projective line X := Z/5 has 6 elements: ∞, 0, 1, 2, −2, −1. Prove that, aspermutations ofX, we have s = (∞, 0)(1,−1) and t = (0, 1, 2,−2,−1). The dodecahedralgroup O is the automorphism group of a dodecahedron. The elements of O permute the 6pairs of opposite faces of the dodecahedron, and so O is a subgroup of S6. Prove that thepermutations s and t are ‘dodecahedral’, and conclude that PSL2(Z/5) = O.

63


64


[Mdlar] 9

2. Small modular groups.

(2.1) Example. Consider the group G := SL2(Z/2). By (1.10)(2), G it is the symmetricgroup S3 of order 6. Hence G has a unique (normal) subgroup G(2) of index 2 (and order3), namely the alternating groupA3 generated by u. The quotientG/G(2) is the cyclic groupC2, and so the canonical map to the quotient may be viewed as a surjective character,

χ2 : G → C2.

In addition,G has three subgroups of index 3 (and order 2), generated by the three involutions,

s =[ 0 1

1 0

], t =

[ 1 10 1

], sts−1 =

[ 1 01 1

].

(Note that 1 = −1 in Z/2.) The latter three subgroups are denoted respectivelyGθ ,G0, andG0.

(2.2) Example. Consider the groupG = SL2(Z/3). It is of order 24 by (1.8). By (1.10)(3),its quotient modulo ±1 is the alternating group A4 of order 12. The alternating group A4contains Klein’s ‘Vier’-group, the kernel of the homomorphism S4 → S3, and the image ofA4 under the homomorphism is the alternating group A3 = C3. Hence, by composition, weobtain a surjective character,

χ3 : G → PSL2(Z/3) = A4 → A3 = C3.

The kernel is denoted G(3). It is of index 3 (and order 8).

(2.3) Example. Consider the group G := SL2(Z/4). It is of order 48 by (1.8). We willdescribe some of its subgroups.

(i) The group G acts on the set X := ((Z/4)2)∗ of columns where at least one coordinateis a unit. There are 12 columns in X, and they are the representatives of the 6 points ofthe projective line Z/4. Hence they also represent the three points ∞, 0, 1 on the projectiveline Z/2. Define a 3-orbit as a subset A = {a, a′, a′′} with 3 elements of X such thata + a′ + a′′ = 0. For instance, the following matrix is a 3-orbit:

[ 1 0 −10 1 −1

]. (1)

In Z/4, the sum of two units is always in the maximal ideal. Hence the sum of any three unitsis a unit. It follows easily for any 3-orbit A = {a, a ′, a′′} that its three columns represent thethree points ∞, 0, 1 in Z/2; consequently, there is a unique ordering of the 3 elements of Asuch that a, a′, a′′ represent respectively ∞, 0, 1. Clearly, the set of all 3-orbits has 42 = 16elements. The groupG acts on the set of 3-orbits. The stabilizer of the 3-orbit (1) consists ofthe matrices in G = SL2(Z/4) whose columns are two of the three columns in (1). Hence,the stabilizer is of order 3. As a consequence, the group G acts transitively on the set of3-orbits.

65


Now, for any given 3-orbit A there is a unique decomposition of X into four 3-orbits,

X = A ∪ B ∪ C ∪D, (2)

with the special property that if E is any 3-orbit in the decomposition, then −E is not in thedecomposition. Indeed, since G acts transitively on the set of 3-orbits, it suffices to provethe assertion when A = {a, a′, a′′} is the 3-orbit (1). Then the unique decomposition is thefollowing:

[ 1 0 −10 1 −1

]∪

[−1 2 −10 −1 1

]∪

[−1 0 12 −1 −1

]∪

[ 1 2 12 1 1

]. (3)

To get the decomposition, take as A the 3-orbit (1). For B = {b, b′, b′′}, take b := −a. Thenb′ �= a′ since a′ ∈ A, and b′ �= −a′ by the special property. Hence the first coordinate of b′is equal to 2. Therefore, the first coordinate of b′′ is equal to −1. As b′′ �= a′′, it follows thatthe second coordinate of b′′ is equal to 1. Therefore, the second coordinate of b′ is equal to−1. Now it is easy to fill in the remaining 6 columns.

Let Y be the set of all decompositions with the special property. It follows that the set Y has4 elements. The groupG acts on the set Y . By the special property, ifD is a decomposition inY , then the composition −D is different fromD. Hence the matrix −1 = s2 acts non-triviallyon Y . Consider the matrix u of order 3. There are 4 decompositions D in the set Y , so atleast one decomposition D is invariant under u. But then also −D is invariant under u, andhence also the two remaining decompositions in the set Y are invariant under u. Thus u actstrivially on Y . SinceG is generated by u and s, it follows that the image of the representation,

G → Aut(Y ) = S4,

is a cyclic group of order 4. Thus the representation may be viewed as a surjective character,

χ4 : G → C4,

where the image is generated by the image of s (or by the image of t , since s = tu). Thekernel of χ4 is a normal subgroupG(4) of order 12 inG. AsG(4) contains all order-3 elementsof G, it is generated by the order-3 elements of G. In particular, G(4) is the unique normalsubgroup of index 4 in G.

(ii) Reduction modulo 2 is a surjective homomorphismG → SL2(Z/2). Hence the kernelof the reduction is a normal subgroup G(2) of order 8. It is easily described: it consists ofthe matrices having two equal units in the diagonal and two non-units off the diagonal. Inparticular, all its elements are of order 2. ThusG(2) it is commutative and elementary abelian(isomorphic to C2 × C2 × C2).

(iii) Consider the intersection,

V := G(4) ∩G(2).

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[Mdlar] 11

It is a normal subgroup of G. It can be described as the kernel of the restriction G(4) →SL2(Z/2) = S3. As G(4) is generated by the order-3 elements of G, the image of therestriction G(4) → S3 is the subgroup A3 of order 3 in S3. Hence V is of order 4 and, beinga subgroup of G(2), V is isomorphic to C2 × C2. Since the decomposition (3) is invariantunder V , it follows that V consists of the following four matrices:

V :[ 1 0

0 1

],

[ −1 20 −1

],

[−1 02 −1

],

[ 1 22 1

]. (4)

(iv) Denote byGθ the subgroup of G consisting of matrices whose reduction modulo 2 isequal to one of the two matrices, [ 1 0

0 1

],

[ 0 11 0

].

In other words, with respect to the reduction map SL2(Z/4) → SL2(Z/2), the subgroupGθis the preimage of the subgroup of the target denotedGθ in Example (2.1). Alternatively,Gθ

can be described as the subgroup ofG stabilizing the following set of 4 columns in ((Z/4)2)∗:[ 11

],

[ 1−1

],

[ −11

],

[−1−1

]. (5)

Clearly, Gθ is a subgroup of index 3 (and order 16), it contains the subgroup G(2) and inaddition the matrix s. It follows that restriction defines a surjective character,

χ4 : Gθ → C4.

The kernel Gθ ∩G(4) is of order 4, and hence equal to V .

(v) The subgroup Gθ has a second character defined as follows: When the four columnsof (5) are multiplied by 2, they become equal, namely equal to the column n whose twocoordinates are equal to 2. Now, there are four subsets {x, x ′} with two elements of X suchthat x + x ′ = n, namely the following four:[ 1

0

][ 12

],

[ 01

][ 21

],

[−10

][−12

],

[ 0−1

][ 2−1

]. (6)

Clearly, the column n is invariant under the matrices ofGθ . Therefore, the groupGθ acts onthe set Z consisting of the four subsets (6). Hence we obtain a representation,

Gθ → Aut(Z) = S4.

Obviously, the matrix s2 = −1 acts nontrivially on Z. Moreover, the set W consisting offollowing four matrices is easily seen to act trivially on Z:

W :[ 1 0

0 1

],

[ 1 20 1

],

[ 1 02 1

],

[ 1 22 1

]. (7)

Since Gθ is of order 16, it follows that the representation is in fact a surjective character,

χθ : Gθ → C4,

and that its kernel is equal to W .

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(2.4) Remark. Let = SL2(Z) be the modular group. Its center is the subgroup ±1.Consider the commutator quotient := / ′. As is generated by the matrices s and u, itfollows that is generated by the images s and u. Moreover, since s4 = u3 = 1, the order ofs is a divisor of 4 and the order of u of order a divisor of 3. Therefore, being commutative,the quotient is cyclic and its order is a divisor of 12.

Now, by Example (2.1), the quotient SL2(Z/2) of has a surjective character onto C2.Hence, by composition, we obtain a surjective character

χ2 : → SL2(Z/2) → C2.

The kernel is denoted (2). Similarly, from Example (2.2), we obtain a surjective character,

χ3 : → SL2(Z/3) → C3,

whose kernel is denoted (3).It follows from the Chinese Remainder Theorem that the intersection (6) := (2) ∩ (3)

is in fact the kernel of the surjective character,

χ6 : → SL2(Z/6) → C2 × C3 = C6.

As a consequence, the order of u is equal to 3 and the order of s is equal to 2 or 4. Clearly,as s2 = −1, the order of s is equal to 4 if and only if −1 /∈ ′. In fact, it follows from themore complicated example (2.3) that there is a surjective character χ4 : → C4. Hence theorder of s is equal to 4 and the commutator quotient is cyclic of order 12.

Therefore, in an obvious notation, the commutator subgroup ′ is equal to (12), where (12) = (4) ∩ (3) is the kernel of a surjective character

χ12 : → C4 × C3 = C12.

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[Mdlar] 13

3. Congruence subgroups.

(3.1) Definition. By Proposition (1.5), reduction modulo N is a surjective homomorphismof groups,

SL2(Z) → SL2(Z/N).

The kernel of the reduction map is denoted (N). Thus (1) is the full modular groupSL2(Z), and (N) is a normal subgroup. A subgroup of (1) containing (N) is called alevel-N (congruence) subgroup. As the reduction map is surjective, the level-N subgroupscorrespond bijectively to the subgroups of SL2(Z/N).

Important series of level-N subgroups are defined by the following congruence conditionsmodulo N on the usual four entries a, b, c, d of a matrix in SL2(Z):

(N) : a ≡ d ≡ ±1, b ≡ c ≡ 0,

0(N) : c ≡ 0,

01(N) : c ≡ 0, d ≡ 1,

00(N) : c ≡ 0, b ≡ 0.

The four subgroups correspond respectively to the following subgroups of SL2(Z/N): thesubgroup ±1, the subgroup of upper triangular matrices, the unipotent subgroup (uppertriangular matrices with 1 in the diagonal), and the subgroup of diagonal matrices. Note thatthe subscripts on the middle two subgroups refer to the second row (c, d) modulo N of thematrix (in the literature, 01 is sometimes denoted 1). Similar congruence subgroups 0(N)

and 10(N) are defined using the first row (a, b).The subgroups (N) for N > 2 are inhomogeneous, that is, they do not contain the

matrix −1. In general, if is a subgroup of SL2(C), we denote by the homogenized group ∪ (− ). Then is a homogeneous group, and the two groups and have the sameimage in PSL2(C), that is, P = P . If is inhomogeneous, then is of index 2 in ,and ∼−→ P . Clearly, the congruence group (N) is the homogenized group of (N).Note that, for N = 2, we have that (2) = (2). The groups 01(N) and 10(N) areinhomogeneous, and they define homogeneous groups 01(N) and 10(N).

(3.2) Observation. Congruence subgroups have finite index in (1) = SL2(Z), becausethe quotient (1)/ (N) is the finite group GN := SL2(Z/N). In fact, the order of GN isdetermined in Proposition (1.8), and it follows that

(1) the index of (N) is equal to N 3 ∏p|N(1 − 1

p2 ).

It is easy to determine the orders of the various subgroups of GN that define the speciallevel-N subgroups of (3.1). It follows that

(2) the index of (N) is equal to 12N

3 ∏p|N(1 − 1

p2 ) when N > 2.

(3) the index of 0(N) is equal to N∏p|N(1 + 1

p).

(4) the index of 01(N) is equal to N 2 ∏p|N(1 − 1

p2 ).

(5) the index of 00(N) is equal to N 2 ∏

p|N(1 + 1p).

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Indeed, the group (N) is the preimage of the subgroup ±1 ofGN , and so the index in (2) isobtained by dividing the order of GN by 2. Similarly, in GN the number of upper triangularmatrices is equal to the product of the orders of Z/N and (Z/N)∗. Hence the index in (3) isobtained by dividing the order ofGN byNϕ(N) = N2 ∏

p|N(1 − 1p). The indices in (4) and

(5) are obtained similarly.

(3.3) Example. Level-2 subgroups correspond to the subgroups ofG= SL2(Z/2) consideredin Example (2.1). The preimage, under the reduction modulo 2, of the subgroup G(2) is anormal subgroup (2) of index 2 in (1). It is equal to the kernel of a surjective character,

χ2 : (1) → C2.

The preimages of the subgroupsG0 andG0 are the subgroups 01(2) = 0(2) and 01(2) = 0(2). In addition, the preimage of the subgroup Gθ is a level-2 subgroup θ , called theθ -group. The latter three subgroups are non-normal subgroups of index 3.

(3.4) Example. The subgroup (3) is of index 24, and the homogeneous group (3) is ofindex 12. The homogeneous level-3 subgroups correspond to the subgroups of the groupPSL2(Z/3) considered in (1.10)(3). In particular, the normal subgroup (3) introduced inRemark (2.4) is a level-3 subgroup of index 3, equal to the kernel of a surjective character,

χ3 : (1) → C3.

(3.5) Example. The subgroup (4) is of index 48. Various level-4 subgroups correspondto the subgroups of G = SL2(Z/4) considered in Example (2.3). The preimage, under thereduction modulo 4, of the subgroupG(4) is a normal subgroup (4) of index 4, equal to thekernel of a surjective character,

χ4 : (1) → C4.

The preimage of G(2) is the level-2 subgroup (2) of index 6. The intersection (4) ∩ (2)is the preimage of V ; it is a normal subgroup of index 12 in (1), and it is denoted V . Notethat (4) and V are non-homogeneous. Their homogenized groups are respectively (2) and (2).

The preimage of the subgroupGθ is the θ -group θ of index 3. It contains the group (2),and in particular the normal subgroup V . Hence restriction defines a surjective character,

χ4 : θ → C4.

In addition, there is a surjective character,

χθ : θ → C4,

whose kernel is the preimage W ofW . Note that W is of index 12 in (1), but not normal.

(3.6) Exercise. Prove that the conjugate group uθ = u−1 θu is equal to 0(2). [Hint:

reduce su = u−1su modulo 2.]

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[Mdlar] 15

(3.7) Exercise. The group (2) is the homogenized group of V . The character χ4 on (1)restricts to a surjective character,

χ4 : (2) → C2.

By example (2.3)(iii), the kernel of the restricted character is equal to V . Prove that therestricted character on (2) is determined by the formula,

χ4

[a b

c d

]= (−1)(a+b+c−1)/2.

[Hint: Clearly, for a matrix in (2), we have modulo 2 that a ≡= 1 and c ≡ d ≡ 0, andhence a + b + c ≡ 1. Therefore, modulo 4 we have that a + b + c ≡ ±1. Now show fromthe description of V in (2.3) that a + b + c ≡ 1 holds for the matrices in V .]

(3.8) Exercise. Clearly, for a matrix in θ we have for the numbers c and d that exactly oneis odd. Prove that the character χθ : θ → C4 is given by the formula,

χθ

[a b

c d

]=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 for d ≡ 1 (mod 4),

i for c ≡ 1 (mod 4),

−1 for d ≡ −1 (mod 4),

−i for c ≡ −1 (mod 4),

where i := χθ(s). [Hint: inspect the cosets of Gθ/W .]

71


72


[Mdlar] 17

4. Some modular fundamental domains.

(4.0) Setup. In this section we describe fundamental domains for some congruence subgroupsof the modular group (1) = SL2(Z) acting on the upper half plane H. The congruencesubgroups are of finite index in (1). Recall that if F is a fundamental domain for a discretegroup and� is a subgroup of finite index in with a given system γi of representatives forthe right cosets of � in , then the union G of the transforms γiF is a fundamental domainfor �. Moreover, if γL is the systems of boundary transformations corresponding to the sideL of F , then a system of boundary transformations of G are obtained as follows: Consider aside of G, say γjL where L is a side of F . Then γLL is a side of F . According to the cosetrepresentation, we have

γjγ−1L = δ−1

j,Lγk

with a unique δj,L ∈ �. Then δj,LγjL = γkγLL is a side of G, and so δj,L is the boundarytransformation corresponding to the side γjL.

(4.1) Example. Take as F the fundamental domain F for the full modular group (1)described in Proposition (Discr.1.6). Each transformγF forγ in (1) is a again a fundamentaldomain, and the transforms cover H.

F

ρ ζ

i

− 12 0 1

2 − 32 −1 − 1

2 0 12 1 3

2

t−1F F tF

uF

u2F

sF

The fundamental domain F for (1), and some of its transforms.

It follows from the description that under the action of (1) there are exactly two ellipticorbits, namely one of order 2 represented by the point i and one of order 3 represented by thepoint ρ. In addition, there is only one parabolic orbit, represented by the point ∞.

Moreover, it follows from the description of the equivalence on the boundary of F that theorbit space H/ (topologically) is a 2-sphere minus a point.

(4.2) Example. Occasionally, to obtain different fundamental domains, it is convenient to cuta domain in subdomains, and then to apply different γ ’s to the different pieces. For instance,F can be cut into two domains, F = F− ∪ F+, defined respectively by the inequalities: z ≥ 0 and z ≤ 0. Then F ′ := t−1F+ ∪ F− is a different fundamental domain for (1)

73


and so is F ′′ := sF+ ∪ F−. For the two latter domains, the boundary transformations aret, u and s, u.

F

F− F+

− 12 0 1

2

F ′

t−1F+ F−

−1 − 12 0

F ′′

F−

sF+

− 12 0 1

2

The fundamental domains F, F ′ and F ′′ for (1).

(4.3) Example. The group (2) is a normal level-2 subgroup of index 2 in (1), equal to thekernel of the character χ2 : (1) → C2. Representatives for the cosets are 1, t . Hence theunion F ∪ tF is a fundamental domain for (2). Similarly, the union F (2) := F ∪ sF is afundamental domain.

F tF

− 12 0 1

2 1 32

F (2)

F

sF

− 12 0 1

2

Fundamental domains for (2).

Consider the first domainF ∪ tF . It has three finite vertices ρ, ζ and ρ+2, and one infinitevertex ∞. Its boundary transformations are t 2 and v := st−1. The angles at the vertices ρand ρ + 1 are equal to 2π/6, the angle at ζ is equal to 2π/3 and the width at ∞ is equal to 1.Hence, there are two elliptic orbits of order 3, represented by ρ and ζ , and one parabolic orbitrepresented by ∞. Clearly, the boundary transformations are t2 and st−1. Topologically, theorbit space H/ (2) is a 2-sphere minus a point.

Similarly, the second domain F (2) has 2 finite vertices ρ and ζ , and two infinite vertices∞ and 0. The angles at ρ and ζ are equal to 2π/3 and the widths at 0 and ∞ are equal to 1

2 .

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[Mdlar] 19

The boundary transformations are defined by the two matrices

u = t−1s =[−1 −1

1 0

], v = st−1 =

[ 0 −11 −1

]. (4.3.1)

As a consequence, modulo ±1, the group (2) is generated by u and v.

(4.4) Example. The group (3) is a normal level-3 subgroup of index 3 in (1), equal to thekernel of the character χ3 : (1) → C3. Representatives for the cosets are the three matrices1, u, and u2. From the fundamental domain F ′ of (4.1), we obtain the fundamental domainG := F ′ ∪ uF ′ ∪ u2F ′ for (3).

G

F ′

uF ′ u2F ′

−1 − 12 0

stG G sG

suG

−2 −1 − 12 0 1

The fundamental domain G for (3), and some of its transforms.

The domain G has no finite vertices, and 3 infinite vertices −1, 0, and ∞. The boundarytransformations correspond to the matrices,

s =[ 0 −1

1 0

], st =

[−1 −21 1

], su =

[−1 −12 1

]. (4.4.1)

It follows that there are 3 elliptic orbits, all of order 2, represented by the points i, i − 1, and− 1

2 + i2 . In addition, the three infinite vertices are equivalent, so there is only one parabolic

orbit. The orbit space H/ (3) is a 2-sphere minus a point.As a consequence, the group (3) is generated by the 3 matrices of (4.4.1).

(4.5) Example. The congruence subgroup 0(2) is a level-2 subgroup of index 3 in (1). Thethree matrices 1, u, u2 represents the right cosets modulo 0(2). So, the domainG of Example(4.4) is also a fundamental domain for 0(2), but, of course, the boundary transformationsare different. For G as a fundamental domain for 0(2), the boundary transformations areassociated to the matrices,

t =[ 1 1

0 1

], su =

[−1 −12 1

]. (4.5.1)

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There is one elliptic orbit, of order 2, represented by the point − 12 + i

2 on the boundary. Itis fixed under su. Of the infinite vertices, −1 and 0 are equivalent. So there are 2 cusps, andH/ 0(2) is a 2-sphere minus 2 points.

As t ∈ 0(2), we can, as in Example (4.2), obtain a different fundamental domain F0(2)by cuttingG by a vertical line and translating the left piece by t .

G

−1 0

F0(2)

− 12 0 1

2

The fundamental domainsG and F0(2) for 0(2).

The domain F0(2) has two finite vertices, ± 12 + i

2 , both of angle 2π/4 and representing(of course) the same elliptic point. In addition, F0(2) has two infinite vertices 0 and ∞. Theboundary transformations of F0(2) correspond to the matrices,

t =[ 1 1

0 1

], sut−1 =

[−1 02 −1

](4.5.2)

As a consequence, the two matrices in (4.5.1) (or, up to ±1 the two matrices in (4.5.2))generate the group 0(2).

(4.6) Example. The θ -group θ is a level-2 subgroup of index 3, and as in the previous twoexamples, the matrices 1, u, u2 represent the right cosets. Hence, again, the domainG is alsoa fundamental domain for θ .

From the decomposition F = F+ ∪F− of Example (4.2), a different fundamental θ canbe obtained as follows: The three matrices 1, t−1, ts form a different set of representativesfor the cosets modulo θ , and so do the three matrices 1, t, t−1s. Apply the first set ofmatrices to the subdomain F+ of F and the second set to the subdomain F−. The resultis a fundamental domain Fθ for θ which is the union of the following 6 pieces: F +,F+

1 := t−1F+, F+2 := tsF+, F−, F−

1 := tF−, and F−2 := t−1sF−. The pieces fit together,

and they form a domain like G with three infinite vertices and no finite vertices.For G as a fundamental domain for θ , the boundary transformations correspond to the

matrices,

s =[ 0 −1

1 0

], st2 =

[ 0 −11 2

]. (4.6.1)

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[Mdlar] 21

There is one elliptic orbit, of order 2, represented by the point i. Of the infinite vertices, ∞and 0 are equivalent. The orbit space H/ θ is a 2-sphere minus 2 points.

G

−1 0

Fθ

F+1 F− F+ F−

1

F−2 F+

2

−1 0 1

The fundamental domainsG and Fθ for θ .

For the fundamental domain Fθ , the boundary transformations are obviously s and t 2. Inparticular, the θ -group θ is generated by the two matrices s and t 2.

(4.7) Example. The group (2) is a normal level-2 subgroup of index 6, equal to the kernelof SL2(Z) → SL2(Z/2) = S3. Coset representatives are the six matrices 1, u, u2, t, tu, tu2.The group (2) is a subgroup of 0(2) and 1, t represent the cosets in 0(2). Hence, withG is the fundamental domain of (4.5), the union F(2) := G ∪ tG is a fundamental domainfor (2). Clearly, the matrix su of (4.3.1) is in 0(2) and not in (2). Hence, the unionH := G ∪ suG is a second fundamental domain for (2).

F(2)

G tG

−1 0 1

H

G

suG

−1 − 12 0

The fundamental domains F(2) and H for (2).

The domain F(2) has no finite vertices; its infinite vertices are −1, 0, 1, ∞. The boundarytransformations are defined by the two matrices,

t2 =[ 1 2

0 1

], st−2s =

[−1 02 −1

]. (4.7.1)

77


Thus, up to ±1, the group (2) is generated by the matrices (4.7.1). Clearly, there are noelliptic points. Of the infinite vertices, −1 and 1 are equivalent. Hence, the orbit spaceH/ (2) is a sphere minus 3 points.

The group (2) is the homogenization of the inhomogeneous level-4 subgroup V de-fined in Example (3.4). Hence the two groups have the same fundamental domain. As aconsequence, V is generated by the two matrices,

−t−2 =[−1 2

0 −1

], st−2s =

[−1 02 −1

]. (4.7.2)

(4.8) Example. The group 0(4) is a level-4 subgroup of index 6. It is a subgroup of 0(2), of index 2. Clearly, the matrix su of (4.3.1) belongs to 0(2) and not to 0(4), andhence it represents the non-trivial coset of 0(4) in 0(2). Hence the fundamental domainH := G∪ suG for (2) described in Example (4.7) is also a fundamental domain for 0(4).Since t is in 0(4), the latter domain can be cut in two by a vertical line, and we can obtain asecond fundamental domain F0(4) by translating the left piece by t .

H

−1 − 12 0

F0(4)

− 12 0 1

2

The fundamental domainsH and F0(4) for 0(4).

The domain H has 4 infinite vertices, ∞, −1, − 12 , and 0. Its boundary transformations

are t and sut (su)−1. For the domain F0(4), the boundary transformations correspond to thematrices,

t =[ 1 1

0 1

], sut−1(su)−1t−1 =

[ −1 04 −1

]. (4.8.1)

In particular, 0(4) modulo ±1 is generated by the two matrices (4.8.1).The orbit space H/ 0(4) is a sphere minus 3 points.

(4.9) Example. The group (6) is a normal level-6 subgroup. It is the kernel of a surjectivecharacter χ6 : SL2(Z) → C6. The cyclic group C6 is generated by the image of t . Thus thepowers t i for i = 0, . . . , 5 are representatives for the cosets, and a fundamental domain Kfor (6) is obtained as the union of the translates t iF for i = 0, . . . , 5.

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[Mdlar] 23

The domain K has ∞ as its only infinite vertex. The finite vertices are the 7 pointswith imaginary part 1

2

√3 and real parts 1

2 j for j = −1, 1, 3, 5, 7, 9, 11. To get the boundarytransformations, note that modulo (6) we have that s ≡ t3. It follows easily that the boundarytransformations are t6, st−3, t4st−1, and t2st−5. Clearly, of the seven finite vertices, the threewith real parts 1

2 , 52 and 9

2 are equivalent and the remaining four are equivalent. The angles inboth classes add up to 2π . Hence, there are no elliptic points for (6). It does require somefeeling for cutting and pasting to see from the domainK that the orbit space H/ (6) is a torusminus one point.

K

F tF t2F t3F t4F t5F

− 12

12

32

52

72

92

112

The fundamental domainK for (6).

A more manageable fundamental domain for (6) is the domain F (6):

F (6)

G sG

−1 0 1

The fundamental domain F (6) for (6).

It is obtained as follows: The level-6 group (6) is contained in the group (3) of Example(4.4), and G was a fundamental domain for the latter group. The boundary transformation sof G is in (3) and not in (6). Hence the two cosets of (6) in (3) are represented by thetwo matrices 1 and t . Hence a different fundamental domain for (6) is obtained as the unionF (6) := G ∪ sG.

79


As sG = tG, the unionF (6) = G∪sG is equal to the fundamental domainF(2) consideredin Example (4.7), but the boundary transformations are different. The domain has 4 infinitevertices, −1, 0, 1, and ∞, and it has no finite vertices.

The boundary transformations correspond to the two matrices,

tst2 =[ 1 1

1 2

], t2st =

[ 2 11 1

]. (4.9.1)

It follows (again) for (6) that there are no elliptic points and exactly one cusp, and theorbit space is a torus minus a point. Modulo ±1, the group (6) is generated by the twomatrices (4.9.1).

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[Autm] 1

Automorphic forms

1. Automorphic factors.

(1.1) Definition. Fix an action of a group G on a set X. As usual, when a second set Y isgiven, the groupG acts on the right on the set Y X of all functions f : X → Y by the definition,

f σ := f σX, (1.1.0)

where σX is the automorphism x �→ σx of X.LetH be a second group. Then anH -valued automorphic factor for the action ofG onX

is a map j : G×X → H satisfying for all σ, τ inG and all x inX the automorphy equation,

j (στ, x) = j (σ, τx)j (τ, x). (1.1.1)

Alternatively, the map j may be viewed as a map σ �→ jσ from G to the group HX of allmaps X → H , and the automorphy equation is the following:

jστ = (jσ τX)jτ . (1.1.2)

Assume thatH acts on the left on the set Y . Then, for everyH -valued automorphic factorj , a right action of G on the set YX of all functions f : X → Y is obtained by the definition,

f ·j σ = j −1σ (f σX), (1.1.3)

or, with arguments,(f ·j σ )(x) = j (σ, x)−1f (σx). (1.1.4)

The action of G on YX is called the automorphic action defined by the factor j . Note that afunction f : X → Y is G-invariant with respect to the automorphic action if and only if, forall x in X and all σ in G,

f (σx) = j (σ, x)f (x). (1.1.5)

(1.2) Note. Clearly, the constant map j (σ, x) = 1 is an automorphic factor. The corre-sponding automorphic action on Y X is given by (1.1.0). More generally, automorphic factorsj (σ, x) that are independent of x correspond to homomorphisms of groups χ : G → H .Functions f : X → Y that are invariant under the automorphic action corresponding to χ areoften called semi invariant. They are characterized by the equation,

f (σx) = χ(σ)f (x).

81

[Autm] 2 Automorphic functions16. februar 1995

(1.3) Note. Assume that the groupG acts on the right on a groupN , by group automorphismsofN . In other words, assume given, for σ inG, a group automorphism ofN denoted n �→ nσ ,such that nστ = (nσ )τ . Then the semi-direct product is the product setG×N with the groupcomposition given by the equation,

(σ, n) · (τ,m) := (στ, nτ m). (1.3.1)

Obviously, the two maps σ �→ (σ, 1) and n �→ (1, n) identify G and N as subgroups ofproduct, and the pair (σ, n) is the product σ · n. Under the identification, the composition(1.3.1) is essentially given by the commutation rule,

n · σ = σ · nσ ,or equivalently, by nσ = σ−1 · n · σ . The projection p : (σ, n) �→ σ is a surjectivehomomorphism of groups p : G × N → G, and its kernel is equal to N . The inclusionσ �→ (σ, 1) is a section of p, that is, it is a homomorphism which composed with p is theidentity ofG. Clearly, the general sections of p are the maps of the form σ �→ (σ, νσ ), whereσ �→ νσ is a 1-cocycle, that is, a map G → N satisfying the following condition,

νστ = (νσ )τ ντ .

Consider, in the setup of (1.1), the semi-direct product G × HX. Then, by (1.1.2), theautomorphic factors G → HX are precisely the 1-cocycles. Equivalently, a map σ �→ jσis an automorphic factor if and only if the map σ �→ (σ, jσ ) is a homomorphism of groupsG → G × HX. As a consequence, an automorphic factor jσ is uniquely determined by itsvalues on a system of generators σj for G.

(1.4) Note. Let j (σ, x) be an H -valued automorphic factor. Clearly, it follows from theautomorphy equation first, that j (1, x) = 1, and next that

j (σ−1, x) = j (σ, σ−1x)−1. (1.4.1)

Assume thatH acts on Y , and consider the corresponding automorphic action ofG on Y X. Asusual, a right action of G is changed into a left action by composing with the anti-involutionσ �→ σ−1 ofG. It follows from (1.4.1) that the corresponding left action ofG on Y X is givenby the equation,

σ ·j f := (jσ f )σ−1X . (1.4.2)

(1.5) Note. In the setup of (1.1), assume that Y is a homogeneous space over H , that is, forany two elements y and y ′ of Y there is a unique element h inH such that y ′ = hy. Clearly,if some function f : X → Y isG-invariant with respect to the automorphic action defined byj , then j is unique, and given by the equation,

j (σ, x) = f (σx)f (x)−1. (1.5.1)

Conversely, whenf : X → Y is any function, then the equation (1.5.1) defines an automorphicfactor j .

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[Autm] 3

(1.6) Lemma. Let α and β be matrices of GL2(C). Let z be a point of C, and assume thatthe two points βz and αβz are different from ∞. Then the following equation holds:

J (αβ, z) = J (α, βz)J (β, z). (1.6.1)

Moreover, if z is a fixed point of β, then the representatives of z are eigenvectors of βcorresponding to the eigenvalue J (β, z).

Proof. The points βz and αβz are represented by the columns

β[z

1

]= J (β, z)

[βz

1

], αβ

[z

1

]= J (αβ, z)

[αβz

1

]. (1.6.2)

It follows that the second equation is obtained by multiplying the first by α. Therefore (1.6.1)holds. Clearly, the last assertion of the Lemma is a consequence of the first equation of(1.6.2).

(1.7) Observation. Consider a finite disk D, that is, a disk not containing the point ∞. ThusD is either the interior of a usual circle in C or the points of an open half plane in C. In theformer case, the point ∞ does not belong to the boundary of D, in the latter case it does.

It follows from (1.6.1) that, for any subgroup of SL(D), the function J (γ, z) is a C∗-valued automorphic factor for the action of on D. More generally, for any integer k, thefunction J (γ, z)k is an automorphic factor.

Note that the absolute value |J (γ, z)|k , for any real number k, is an automorphic factor.Our principal interest is, for certain subgroups , automorphic factors j such that j (γ, z) is,at least up to a complex sign, a determination of the function J (γ, z)k where the exponent kis an arbitrary real number.

(1.8) Remark. Fix a real number k. Let w be a non-zero complex number. Recall that acomplex number u is called a value of wk , if |u| = |w|k and if, for some argument θ of w,the real number kθ is an argument for u. If k is integral, there is only one value of wk , if k isrational there is a finite number of values, and if k is irrational there is an infinite number ofvalues.

Let g : X → C∗ be a continuous map. Recall that a map h : X → C∗ is called a deter-mination of gk , if h is continuous and, for every x in X, h(x) is a value of g(x)k . If h0 is adetermination of gk then, for every integer n, the following product is a determination of gk:

e2πinkh0(x); (*)

moreover, if X is connected, then all other determinations are of the form (*).Determinations of gk exist if X is simply connected, or if the image of g(X) is contained

in a simply connected subset of C∗. In the latter case there exists a continuous argumenton g(X), that is, a real valued continuous function � defined on g(X) such that �(w) is anargument for w for all w in g(X), and then the following function is a determination,

h(x) := |g(x)|kei�(g(x))k,

83


In most of our applications, the map g will be holomorphic, and the image of g will avoid ahalf ray starting at the point 0. In this case, the existence of a continuous argument is obvious,and any determination of gk is again a holomorphic function. For example, if the imageg(X) avoids the negative real axis, we may always consider the principal determination ofgk , defined using as argument of g(x) the principal argument θ , where −π < θ < π .

For instance, let D be a finite disk and let α be a matrix in GL2(C) such that αz �= ∞for all z in D. Then the function J (α, z) is (finite) and non-zero everywhere in D. It has theform Cz + D. If C = 0, then the function J (α, z) is the non-zero constant D. If C �= 0,then J (α, z) is a Möbius transformation, and in particular, the image of D is finite disk notcontaining the point 0. Therefore, in both cases, there are determinations of the functionJ (α, z)k. In the first case, any determination is constant, and hence defined on all of C. Inthe second case, the function J (α, z) has a zero v outside D. Let V be the open subset ofC obtained by cutting away a half ray starting at v and having no other points in commonwith the boundary of D. Then there is a determination of J (α, z)k defined on all of V . Inparticular, there is a determination J (α, z)k defined on D, and it extends continuously to thepoints of ∂D except possibly to the two points ∞ and v = α−1∞ if they belong to ∂D.

(1.9) Definition. Let D be a finite disk, and let be a subgroup of SL(D). A factor of weightk on is a function j (γ, z) defined for γ ∈ and z ∈ D and satisfying the following threeconditions:

(1) The function j is a C∗-valued automorphic factor for the action of on D.(2) For fixed γ in , the function j (γ, z) is holomorphic.(3) The absolute value |j | is equal to |J |k , that is, for all γ and z,

|j (γ, z)| = |J (γ, z)|k.

The group C∗ acts on itself, it acts on C, and it acts on C. In particular, from a given factorj on we obtain a corresponding automorphic action of on the numeric functions on D.The action is given by the formula,

(f ·j γ )(z) = 1

j (γ, z)f (γ z). (1.9.1)

Numeric functions on D that are invariant under the action are called ( , j)-invariant, orj -invariant. They are characterized by the equations, for γ ∈ and z ∈ D,

f (γ z) = j (γ, z)f (z). (1.9.2)

Obviously, the action (1.9.1) is linear. Hence, the j -invariant functions form a vector spaceover C.

(1.10) Lemma. Let D be a finite disk, and a subgroup of SL(D). Assume that j is a factorof weight k on . Fix a matrix γ in , and consider a determination J (γ, z)k on D. Thenthere is a complex sign ε (i.e., |ε| = 1), and an equation,

j (γ, z) = εJ (z, γ )k for all z ∈ D, (1.10.1)

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[Autm] 5

In particular, the function j (γ, z) is completely determined by any of its values j (γ, z0) (andthe given weight k).

Proof. By Condition (1.9)(3), the quotient j (γ, z)/J (γ, z)k is of modulus 1, and holomorphicby Condition (2). Therefore, the quotient is a constant function, and hence equation (1.10.1)holds.

(1.11) Observation. (1) The constant automorphic factor j (γ, z) = 1 is a factor of weight 0,and the corresponding invariant functions are simply -invariant functions: f (γ z) = f (z)

for all γ in .More generally, any factor j (γ, z) of weight 0 is of constant modulus 1, and hence constant.

Therefore, the factors of weight 0 on are precisely the unitary characters χ : → C∗.(2) Clearly, the product j1j2 of factors of weights k1 and k2 is a factor of weight k1 + k2.

If f1 is j1-invariant and f2 is j2-invariant, then the product function f1f2 is j1j2-invariant.Similarly, the quotient j1/j2 is a factor of weight k1 − k2. As a consequence, if j0(γ, z)

is given factor of weight k, then the factors of weight k are precisely the functions,

j (γ, z) = χ(γ )j0(γ, z),

where χ : → C∗ is a unitary character. In particular, if the commutator subgroup ′ is offinite index in , then the number of factors of weight k is either zero or equal to the index| : ′|.

(3) If k is an integer, then the function J (γ, z)k is a factor of weight k. In particular, itfollows from (2) that the factors of integral weight k on are the functions,

χ(γ )J (γ, z)k, (1.11.1)

where χ : → C∗ is a unitary character. For an integer k, the action (1.9.1) of on numericfunctions corresponding to the factor J (γ, z)k is denoted f ·k γ , that is,

(f ·k γ )(z) = 1

J (γ, z)kf (γ z).

Numeric functions on D that are invariant with respect to the factor J (γ, z)k are said to be -invariant of weight k.

(4) As noted in (1.4), it follows from the automorphy equations that j (1, z) = 1. Assumethat is homogeneous, that is, −1 ∈ . Then it follows similarly that j (−1, z)2 = 1.Consequently, the function j (−1, z) is the constant 1 or −1. If a function f is j -invariant,and f is non-zero, say f (z0) �= 0, then it follows from the equation f (z0) = f ((−1)z0) =j (−1, z0)f (z0) that j (−1, z0) = 1. Hence j (−1, z) is the constant 1. Therefore, when ishomogeneous, we are mostly interested in homogeneous factors, that is, factors j for whichj (−1, z) = 1.

Note that the factor J (γ, z)k for an integer k is only homogeneous when k is even. Thehomogeneous factors of odd (integral) weight k are the functions (1.11.1) where χ : → C∗

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is an odd unitary character (that is, χ(−1) = −1). In particular, if −1 belongs to thecommutator subgroup ′, then there are no homogeneous factors of odd integral weight on .

(5) Clearly, if is inhomogeneous, and denotes the homogenized group = ∪ (− ),then every factor j on extends to a homogeneous factor j on by defining j (−γ, z) :=j (γ, z) for γ in .

(6) The function |J (γ, z)|k is an automorphic factor on , but it is not holomorphic, andhence it is not a factor in the sense of Definition (1.9). Note that the condition (1.9)(3) is thatthe automorphic factor |j (γ, z)| is equal to the automorphic factor |J (γ, z)|k . Clearly, for anautomorphic factor j (γ, z) to satisfy condition (1.9)(3), it suffices that the equations,

|j (γ, z)| = |J (γ, z)|k, (1.11.2)

hold for a system of matrices γ generating the group .(7) As noted in (1.5), if f : D → C∗ is any function, then the equation,

j (γ, z) := f (γ z)/f (z), (1.11.3)

defines an automorphic factor on any subgroup of SL(D). If f is holomorphic, then j isholomorphic in z. In fact, if f is meromorphic and the right hand side of (1.11.3) has nozeros or poles, then (1.11.3) defines a holomorphic automorphic factor j . But of course, forj to be a factor in the sense of Definition (1.9) it is required that the equations,

|f (γ z)/f (z)| = |J (γ, z)|k, (1.11.4)

hold for all matrices γ in the group . As observed in (6), it suffices that the equations(1.11.4) hold for a system of generators for .

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[Autm] 7

2. Examples I.

(2.1) Example. Let k be an integer. Let � be a lattice in C, say � = Zω1 + Zω2 where thecomplex numbers ω1 and ω2 are linearly independent over R. Consider the following sum,

Ek(�) = Ek

[ω1ω2

]=

∑ω �=0

1

ωk, (2.1.1)

where the sum is over all ω �= 0 in �. The sum is absolutely convergent for k ≥ 3. Fortypographical reasons, the sum will also be denoted E(ω1, ω2). In particular, for z ∈ H, wedefine the Eisenstein series,

Ek(z) = Ek(z, 1) =∑′ 1

(nz +m)k, (2.1.2)

where the sum is over pairs of integers (m, n) �= (0, 0). The function Ek(z), for k ≥ 3, isholomorphic in H.

Clearly, if λ is a non-zero complex number, then Ek(λ�) = λ−kEk(�). In addition, ifγ is a matrix in (1) = SL2(Z) then the two pairs (ω1, ω2) and (ω1, ω2)γ

tr generate thesame lattice. As a consequence, E(ω1, ω2) = E(ω1, ω2)γ

tr. In particular, we obtain theequations,

Ek(γ z) = Ek

[γ z

1

]= Ek

1

J (γ, z)γ[z

1

]= J (γ, z)kEk(z).

In other words, for an integer k ≥ 3, the function Ek(z) is (1)-invariant of weight k.Note that the function Ek(z) is equal to zero when k is odd. We will see later that the

function in non-zero when k is even.

(2.2) Example. The Eisenstein series (2.1.2) is closely related to the following series, definedfor an integer k ≥ 3,

Gk(z) := 1

2

∑(m,n)=1

1

(mz + n)k, (2.2.1)

where the sum is over all pairs of relatively prime integers (m, n). Indeed, if we group in(2.1.2) the terms corresponding to the greatest common divisor d of (m, n), we obtain theequation,

Ek(z) = 2ζ(k)Gk(z), (2.2.2)

where ζ(k) = ∑d≥1 d

−k . Hence, the function Gk is (1)-invariant of weight k. It vanisheswhen k is odd. If k is even, then the value ζ(k) is a well known rational multiple of π k . Moreprecisely, then 2ζ(k) = −(2πi)kBk/k! where Bk is the k’th Bernoulli number.

(2.3) Example. Dedekind’s η-function is the function in the upper half plane H defined bythe product,

η(z) = e2πiz/24∏n≥1

(1 − e2πinz). (2.3.1)

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The sum∑

n e2πinz converges normally in H; hence, so does the product. Consequently, the

η-function is holomorphic and everywhere non-zero in H.The η-function satisfies the following functional equations,

η(z + 1) = e2πi/24η(z), η(−1/z) =√z/i η(z), (2.3.2)

where√z/i is the principal determination. The first equation is obvious, but the second is

far from trivial. It follows from the functional equations that

|η(z + 1)| = |η(z)|, |η(−1/z)| = |z| 12 |η(z)|.

The modular group (1) = SL2(Z) is generated by the two matrices t and s, and tz = z+ 1and sz = −1/z. Moreover, J (t, z) = 1 and J (s, z) = z. Hence, the equation,

|η(γ z)/η(z)| = |J (γ, z)| 12 ,

holds for the two generators s and t . It follows that the equation holds for all matrices in (1).Therefore, as observed (1.11)(7), a factor jη of weight 1

2 on (1) is defined by the equation,

jη(γ, z) = η(γ z)/η(z). (2.3.3)

Of course, the η-factor jη is fully determined by the automorphy equations from the specialvalues,

jη(t, z) = e2πi/24, jη(s, z) =√z

i,

but a priori, it is far from obvious that these two values define an automorphic factor on (1).By construction, the function η is jη-invariant. As the weight of jη is 1

2 , the even powersj 2kη are of integral weight k, and hence of the form in (1.11.1). In particular, the square j 2

η isa factor of weight 1 on (1), of the form,

j 2η (γ, z) = χ(γ )J (γ, z),

where χ : (1) → C∗ is a unitary character. For γ := t we obtain the equation e2πi/12 =χ(t). For γ = s we obtain the equation z/i = χ(s)z, that is, χ(s) = e−2πi/4. As t ands generate (1), it follows that χ maps into the cyclic subgroup C12 of C∗. A priori, thecharacter group of (1) is cyclic of order a divisor of 12, since (1) is generated by thematrices s and u of orders 4 and 3. Hence, using the functional equation of the η-function,we recover the result of (Mdlar.2.4) that the character group is cyclic of order 12.

As a consequence, the power jη(γ, z)2k , where k is an integer divisible by 12, is equal toJ (γ, z)k . In particular, the discriminant�(z), defined by the equation,

�(z) := η(z)24,

is a (1)-invariant function of weight 12 on H. Like η(z), the discriminant is everywherenon-zero.

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[Autm] 9

(2.4) Example. As the η-function is everywhere non-zero in H, there is, for an arbitrary realnumber k, a determination of the function η(z)2k . In fact, a canonical determination of η2k

is selected by chosing for the first factor in the product (2.3.1) the determination e2πikz/12

and for the n’th factor in the product the principal determination (1 − e2πinz)2k (given by thebinomial expansion). It follows that a factor of weight k on (1) is defined by the equation,

j2kη (γ, z) := η(γ z)2k/η(z)2k.

It follows in particular that for any given real number k there are exactly 12 factors of weightk on (1).

(2.5) Example. The function G4 of Example (2.2) is (1)-invariant of weight 4. Hence thecube G 3

4 is of weight 12. The discriminant� is of weight 12, and it is everywhere non-zero.Therefore, the following function,

j (z) := G4(z)3/�(z), (2.5.1)

is of weight 0, that is, the function j (z) is a (1)-invariant function in the usual sense:j (γ z) = j (z) for all matrices γ in (1). The function j is called Klein’s j -invariant. Weprove later the following equation,

123�(z) = G4(z)3 −G6(z)

2. (2.5.2)

It allows an expression of j (z) in terms of G4 and G6. In terms of the Eisenstein series, it iscustomary to write,

g2(z) := 60E4(z), g3(z) = 140E6(z).

Then, using the values of the Bernoulli numbers, B4 = −1/30 and B6 = 1/42, it is easy toderive the equation,

j (z) = 123 g2(z)3

g2(z)3 − 27g3(z)2. (2.5.3)

We show later that Klein’s invariant defines an isomorphism,

H/ (1) ∼−→ C,

from the orbit space to the Riemann sphere.

(2.6) Example. The θ -function is the function defined in the upper half plane H as the sumover all integers n,

θ(z) =∑

eπin2z. (2.6.1)

The θ -function is holomorphic in H and it satisfies the two functional equations,

θ(z + 2) = θ(z), θ(−1/z) =√z/i θ(z), (2.6.2)

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where√z/i is the principal determination. As in Example (2.3), the first equation follows

immediately from the definition, but the second is far from trivial. In fact, the second equationis equivalent to the functional equation for Riemann’s ζ -function. It follows from the twoequations that |θ(z + 2)| = |θ(z)| and |θ(−1/z)| = |z|1/2|θ(z)|.

The Möbius transformations z �→ z + 2 and z �→ −1/z are associated to the matrices t 2

and s. Moreover, J (t2, z) = 1 and J (s, z) = z. Therefore, the two equations imply that theθ -function satisfies the equation

|θ(γ z)/θ(z)| = |J (γ, z)| 12 ,

for the two matrices γ = t2 and γ = s. The latter two matrices generate the θ -group θ by(Mdlar.4.6). It follows that the θ -function determines a unique factor jθ of weight 1

2 on thegroup θ , given by the equation,

jθ (γ, z) = θ(γ z)/θ(z). (2.6.3)

The θ -factor jθ is of course fully determined by its values on the two generators,

jθ (t2, z) = 1, jθ (s, z) =

√z/i,

but a priori, it is far from obvious that these two values define an automorphic factor on θ .

(2.7) Remark. The sum (2.6.1) defining the θ -function can be split into two parts, θ =θev + θodd, where the sums are respectively over the even and odd integers. Obviously,θev(z) = θ(4z), and hence θodd(z) = θ(z) − θ(4z). Moreover, θev(z + 1) = θev(z) andθodd(z+ 1) = −θodd(z). Hence we obtain the functional equation,

θ(tz) = 2θ(4z)− θ(z). (2.7.1)

From the functional equation (2.7.1) for θ(tz) and the functional equation (2.6.2) for θ(sz)we can obtain a functional equation for θ(γ z) for any matrix γ in (1). Consider for instancethe matrix u = t−1s defining the Möbius transformation uz = −1 − 1/z. From (2.7.1) and(2.6.2) we obtain the equations,

θ(−1 − 1/z) = θ(−1/z+ 1) = 2θ(−4/z)− θ(−1/z) = 2√z/4i θ(z/4)−

√z/iθ(z).

Hence,θ(uz) =

√z/i

(θ(z/4)− θ(z)

). (2.7.2)

It follows easily from (2.7.1) and (2.7.2) that the two functions, θ(tz) and θ(uz), are givenby the sums,

θ(z+ 1) =∑n

(−1)neπin2z, θ(−1 − 1/z) =

√z

i

∑n odd

eπin2z/4. (2.7.3)

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[Autm] 11

3. The signs of a factor.

(3.1) Setup. Fix a finite disk D and a discrete subgroup of SL(D). In addition, fix a realnumber k and a factor j (γ, z) of weight k on .

(3.2) Lemma. Let γ be a matrix in , and let u be a fixed point of γ in the closure of D.Then the following limit exists:

j (γ, u) := limz→u

j (γ, z), (3.2.1)

where the limit is taken over points z of D. If γ is elliptic or equal to ±1, then j (γ, u) is ad’th root of unity where d is the order of γ . If γ is parabolic, then |j (γ, u)| = 1.

Proof. Consider a determination of J (γ, z)k on D. By Lemma (1.10), there is an equation,

j (γ, z) = εJ (γ, z)k,

where |ε| = 1. As remarked in (1.8), it follows that the function j (γ, z) extends continuouslyto the points of ∂D, except possibly to the points ∞ and γ −1∞. Since u is a fixed point, itis only exceptional when it is equal to ∞. In the exceptional case, the function J (γ, z) isconstant. Therefore j (γ, z) is constant and hence it extends trivially to all of C. Thus, in allcases, the limit (3.2.1) is simply the value at u of the extended function j (γ, z).

The remaining assertions hold trivially if γ = ±1. Assume next that γ is elliptic. Then thefixed point u belongs to D. Clearly, since u is a fixed point, it follows from the automorphyequation that j (γ, u)d = j (γ d, u) = 1. Hence j (γ, u) is a d’th root of unity.

Assume next that γ is parabolic. Then the fixed point u belongs to the boundary ∂D. Ifu �= ∞, then by (1.6), the number J (γ, u) is an eigenvalue of γ , and hence equal to ±1since γ is parabolic. Consequently, if u �= ∞ then j (γ, u) = εJ (γ, u)k is of modulus 1.Clearly, if the fixed point u is equal to ∞, then the function J (γ, z) is constant, and, since γis parabolic, it is the constant ±1. Hence, j (γ, z) is a constant of modulus 1. In particular,the limit j (γ, u) is of modulus equal to 1.

Thus the assertions have been proved in all cases.

(3.3) Definition. Recall that for any point u in D ∪ ∂ D we have defined the canonicalgenerator γu of at u. It belongs to the isotropy group u. It follows from Lemma (3.2) thatthe following number,

ωu = ωu(j, ) := j (γu, u),

is of modulus 1. It is called the sign of the factor j at the point u. The unique real numberκu = κu(j) determined by the conditions,

ωu = e2πiκu , 0 ≤ κu < 1,

is called the parameter of the factor j at the point u.

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At a -ordinary point u, the canonical generator is equal to 1. Hence ωu = 1 and κu = 0.Assume that u is a -elliptic point of D. Then, by (3.2), the sign is a du’th root of unity,where du is the order of γu. Hence the parameter κu is of the form κu = a/du where a is aninteger and 0 ≤ a < du. In particular, we obtain for any point u of D the inequalities,

0 ≤ κu ≤ 1 − 1/du. (3.3.1)

Clearly, if −1 ∈ and the factor j is homogeneous, then the sign ωu is an eu’th root of unity,where eu = |P u| is the order of the -elliptic point u, and in the inequalities we can replacedu by eu.

(3.4) Proposition. Assume that is the modular group (1) = SL2(Z) acting on the upperhalf plane H. Then the sign of the factor j at the point ∞ is of the form,

ω∞ = ζ−1e2πik/12, (3.4.1)

where ζ is a 12th root of unity. Moreover, the factor j is completely determined by the rootζ and the given weight k. At the elliptic points i and ρ, the signs of the factor j are thefollowing:

ωi = ζ 3, ωρ = ζ 4.

Finally, j (−1, z) = ζ 6 for all z.

Proof. The point ∞ is parabolic, and the matrix t is the canonical generator γ∞. As J (t, z) =1, it follows that j (t, z) is constant, and hence we obtain the equation,

j (t, z) = ω∞.

The point i is elliptic, and the matrix s of order 4 is the canonical generator γi . As J (s, z) = z,the function j (s, z) is, by Lemma (1.10), of the form εzk where |ε| = 1 and zk is the principaldetermination in H. Evaluation at i yields ωi = εik . Hence,

j (s, z) = ωizk/ik,

and the sign ωi is a 4th root of unity. Now s = tu. Therefore, from the automorphy equation,we obtain that

j (u, z) = j (s, z)/j (t, uz) = ω−1∞ ωiz

k/ik.

The point ρ is elliptic, and the matrix u of order 3 is the canonical generator γρ . Hence thesign ωρ = j (u, ρ) is a 3rd root of unity. Take z = ρ in the expression for j (u, z) to obtainthe equation,

ω∞ = ωi/ωρ · ρk/ik.Clearly, ρk/ik = e2πik/12 and ζ := ωρ/ωi is a 12th root of unity. Hence ω∞ has theform (3.4.1). Moreover, the assertions about ωi and ωρ follow from the definition of ζ .Furthermore, j is uniquely determined, because is generated by s and t , and j (t, z) andj (s, z) were determined above by the signs at ∞ and i. Finally, j (−1, z) is constant; theconstant is equal to j (−1, i) = j (s2, i) = j (s, i)2 = ω2

i = ζ 6.

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[Autm] 13

(3.5) Note. In (3.4), there are twelve possible values for ζ , and hence at most 12 possiblefactors of a given weight k. As noted in example (2.4), there are in fact 12 factors of anygiven real weight k.

Clearly, for an integral weight k, the factor J (γ, z)k corresponds to ζ = e2πik/12.

(3.6) Example. The factor jη associated with Dedekind’s η-function is a factor of weightk = 1

2 on the modular group (1). As jη(t, z) = e2πi/24, we have, in the notation of (3.4),that ζ = 1. Hence we obtain for jη the following signs,

ω∞ = e2πi/24, ωi = 1, ωρ = 1.

The corresponding parameters are 1/24, 0, and 0.The square j 2

η is one of the possible 12 factors of weight 1 on (1). In fact, by (2.3),

j 2η (γ, z) = χ12(γ )J (γ, z),

where χ12 : (1) → C12 is the unique character for which χ12(t) = e2πi/12. As ω∞ =e2πi/12, we have ζ = 1 in (2.4.1), and we obtain for j 2

η the following signs:

ω∞ = e2πi/12, ωi = 1, ωρ = 1.

(3.7) Example. The θ -group θ ⊂ SL2(Z), acting on the upper half plane H, has one ellipticorbit represented by the point i, and two cusps represented by the points ∞ and −1. At thethree points, the canonical generators are the matrices,

γi = s =[ 0 −1

1 0

], γ∞ = t2 =

[ 1 20 1

], γ−1 = utu−1 =

[ 2 1−1 0

].

Indeed, the first two equations are obvious. To find the canonical generator at −1, applyconjugation by u. Under the conjugation, the point ∞ corresponds to u(∞) = −1 and θcorresponds to the conjugate group u

θ = u−1 θu. Modulo 2, we have that u−1su ≡ t , andit follows that the conjugate group u

θ is equal to 0(2), cf. Exercise (Mdlar.3.6). Clearly,for the conjugate group the canonical generator at ∞ is the matrix t . Thus γ−1 = utu−1.

Consider the θ -factor jθ on θ . It is of weight 12 and determined by the equations of (2.6),

jθ (γ∞, z) = 1, jθ (γi, z) =√z/i.

Clearly, γ−1 = −γ −1∞ γi . Therefore, from the automorphy equation, we obtain that

jθ (γ−1, z) = jθ (γ−1

∞ , γiz)j (γi, z) = 1 ·√z/i =

√z/i.

As√z/i → e2πi/8 for z → −1, we obtain for jθ the following signs,

ω∞ = 1, ω−1 = e2πi/8, ωi = 1.

The corresponding parameters are 0, 1/8, and 0.

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(3.8) Lemma. Let L be a linear function, L(z) = Cz + D, where C �= 0. Assume thatL(z) �= 0 everywhere on D, and consider a determination of L(z)k on D. Let γ be a matrixin SL(D) and let u be a fixed point of γ belonging to the closure of D. If u belongs to theboundary of D, assume moreover that γ is parabolic. Then,

limz→u

L(γ z)k/L(z)k = 1, (3.8.1)

where the limit is taken over points z of D.

Proof. The functionL has one zero v = −D/C in C. Let V be the open subset of C obtainedby cutting away from C a half ray starting at v and having no other points in common withthe boundary of D. Then D is contained in V , and clearly there is a unique extension ofL(z)k to a determination defined on all of V . Therefore the assertion (3.8.1) is elementarywhen u belongs to V : in the fraction, both the denominator and the numerator converges tothe non-zero value L(u)k .

It remains to consider two further possible cases: u = ∞ and u = v. Clearly, to prove(3.8.1), it suffices to prove that the quotient,

L(γ z)/L(z), (3.8.2)

converges to 1 for z → u. The map L is a Möbius transformation. Therefore, after aconjugation, replacing D byL(D), by L L−1, and γ byLγL−1, we may assume thatL isthe identity,L(z) = z. Thus D is a finite disk not containing the point 0, and the two possiblecases are u = ∞ and u = 0.

Assume first that u = ∞. Then D is a half plane and, since γ is parabolic, the associatedtransformation is of the form γ z = z + b. Then the fraction (3.8.2) is equal to 1 + b/z, andit converges to 1 for z → ∞. Assume next that u = 0. Then D is a disk containing 0 on itsboundary. Since γ is parabolic with 0 as fixed point, the associated transformation is of theform γ z = z/(cz + 1). Then the fraction in (3.8.2) is equal to 1/(cz + 1), and it convergesto 1 for z → 0.

Thus the assertion has been proved in all cases.

(3.9) Exercise. In the setup of (3.8), assume that the fixed point u is on the boundary of D,but assume that γ is hyperbolic. Prove the limit in (3.8.1) exists, and find its value.

(3.10) Definition. Let α be a matrix of GL2(C) mapping a finite disk D′ onto the finite diskD. Let ′ := α = α−1 α be the conjugate subgroup. It is a discrete subgroup of SL(D′).Consider a determination J (α, z′)k on D′. Define the conjugate jα of j as the function on ′ × D′ given by the equation,

jα(γ ′, z′) := J (α, z′)k

J (α, γ ′z′)kj (αγ ′α−1, αz′). (3.10.1)

Clearly, the fraction on the right is independent of the choice of determination.Note that the conjugate factor is not obtained simply by a transport of structure. The latter

transport would yield the function j (αγ ′α−1, αz′).

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[Autm] 15

(3.11) Lemma. In the setup of (3.10), the conjugate j ′ := jα is a factor of weight k onthe conjugate group ′ := α . Moreover, if u′ and u = αu′ are corresponding points ofD′ ∪ ∂ ′D′ and D ∪ ∂ D, then the signs are equal:

ωu′(j ′) = ωu(j). (3.11.1)

If the matrix α belongs to , then ′ = and j ′ = j . In particular, the signs of j at two -equivalent points are equal.

Proof. In (3.10.1), the function j (αγ ′α−1, αz′) is an automorphic factor for the action of ′onD′, because it is obtained by a simple transport of structure. Moreover, as noted in (1.5), thefraction in (3.10.1) is an automorphic factor. Therefore, the function j α is an automorphicfactor on ′. Clearly, the function jα(γ ′, z′) is holomorphic in z′. Finally, the condition(1.9)(3) for jα follows by applying (1.6) to αγ ′ = γα, where γ := αγ ′α−1. Hence theconjugate factor jα is a factor of weight k.

Assume that u = αu′. Then, if γu is the canonical generator of u, the conjugate matrixα−1γuα is the canonical generator γu′ of ′

u′ . Therefore, the equality (3.11.1) follows fromLemma (3.8).

Assume that α belongs to . Then ′ = . It follows from Lemma (1.11) that in thedefinition of jα in (3.10.1), we may replace the fraction J (α, z)k/J (α, γ z)k by the fractionj (α, z)/j (α, γ z). Then the equation jα = j follows by applying the automorphy equationto αγ ′ = γα, where γ =: αγ ′α−1.

Clearly, the last assertion of the Lemma is a consequence of (3.11.1).

(3.12) Note. (1) For an integral weight k, the factor J (γ, z)k is invariant under conjugation,as it follows from (1.3).

(2) In the setup of (3.10), let β be a matrix mapping a finite disk D′′ onto D′. Thenjαβ = (jα)β , as it follows by an easy computation.

(3) It follows from the last part of Lemma (3.11) that the signs ωu of j are completelydetermined by the signs at one point in each parabolic or elliptic orbit.

(4) Clearly, if j1 and j2 are factors of weights k1 and k2 on , then j1j2 is a factor ofweight k1 + k2, and for the signs we have the equation ωu(j1j2) = ωu(j1)ωu(j2).

(5) The signs ωu( , j) depend on the group , that is, they change if j is restricted to asubgroup � of . Assume that � is of finite index in . Then ∂�D = ∂ D. Consider apoint u in D ∪ ∂ D. Denote by d the index d := |P u:P�u|. Assume for simplicity that �is homogeneous and that the factor j is homogeneous. If γu is the canonical generator of u,then γ du is the canonical generator of �u. Therefore we obtain the equation,

ωu(�, j) = ωu( , j)d .

(3.13) Proposition. For j (γ, z) = J (γ, z)k where k is an integer, the sign is equal to 1 at aregular cusp, it is equal to (−1)k at an irregular cusp, and at a point u of D it is equal toeπik(1−1/eu) where eu = |P u|. In particular, for k = 2, the parameter is equal to 0 at anycusp, and equal to 1 − 1/eu at a point u ∈ D.

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Proof. For a cusp u we may, by conjugation, assume that (D, u) = (H,∞). Hence thecanonical generator γu is equal to th is the regular case and equal to −th in the irregular case.Therefore, in the irregular case, we have that J (γu, z) is the constant −1; it follows that signat u of J k is equal to (−1)k . In particular, the parameter is equal to 0 when k is even, andequal to 1

2 when k is odd. The assertions in the regular case are proved similarly.For a point u of D, we may, by conjugation, assume that (D, u) = (E, 0). Then the

canonical generator γu is the matrix of (Discr.1.3.1) with d := eu. Thus J (γu, z) is theconstant −e−πi/eu = eπi(1−1/eu). It follows that the sign of J k at u is equal to eπik(1−1/eu).In particular, the parameter is the fractional part of k

2 (1 − 1/eu). Thus, for k = 2, theparameter is equal to 1 − 1/eu.

(3.14) Example. Assume that j is a factor on (1), determined as in Proposition (3.4) by itsweight k and a 12th root of unity ζ . Restrict j to the subgroup θ . At the point i, the canonicalgenerators are equal and so we obtain the equation ωi( θ , j) = ωi( (1), j). At the point ∞,the canonical generator of θ is equal to t2, and hence ω∞( θ , j) = ω∞( (1), j)2. Finally,at the point −1, the two groups have the same canonical generator, and hence ω−1( θ , j) =ω−1( (1), j) = ω∞( (1), j). Hence, by the results of Proposition (3.4), we obtain for therestriction of j to θ the following signs:

ω∞ = ζ−2e2πik/6, ω−1 = ζ−1e2πik/12, ωi = ζ 3.

In particular, the factor jθ of Example (3.7) is not the restriction of a factor on (1).

(3.15) Exercise. On the θ -group θ , the two factors jη and jθ differ by a unitary characterχ : θ → C∗. Identify the character.

(3.16) Definition. Assume that the matrix α in SL2(C)maps the finite disk D onto the finitedisk D′. Consider a determination J (α, z′)k on D′ and a complex sign ε. For any numericfunction f on D, we define the weight-k conjugate to f as the function,

f α(z) := ε

J (α, z′)kf (αz′). (3.16.1)

The definition of a weight-k conjugate function does not depend on the given subgroup of SL(D) and in particular, it is independent of the factor j . However, it should be notedthat contrary to the definition of the conjugate factor j α , the definition of the conjugatefunction is ambiguous: f α depends on the choice of determination J (α, z′)k . As differentdeterminations differ by a complex sign there is, strictly speaking, only a well defined classof conjugate functions that differ by a complex sign.

Clearly, a function f is ( , j)-invariant if and only if the weight-k conjugate function f α

is ( α, jα)-invariant. Note also that if α belongs to (in which case D′ = D), then f ·j αis a weight-k conjugate to f .

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[Autm] 17

4. Automorphic forms.

(4.1) Definition. A numeric function f (with values in C) defined on the upper half plane Hwill be called exponentially bounded at ∞, if there exists a real number C such that

f (z) = O(|eiz|C) for z → ∞, (4.1.1)

uniformly on any vertical strip in H. The positive function |eiz|C is equal to e−Cz and thecondition on C implied by the O-notation is equivalent to the following: For every verticalstrip in H, there are positive real numbers R and M so that the inequality,

|f (z)| ≤ Me−Cz,

holds when z is in the strip and z ≥ R.In the most important of the cases to be considered, the function f will in fact be periodic,

with a real period. Clearly, in this case, the condition holds for C if and only if (4.1.1) holdsuniformly on all of H.

The supremum of the numbersC for which the condition holds, will be called the order off at ∞ and it will be denoted ordH∞ f . The number ordH∞ f is equal to +∞ if the conditionholds for all C, and equal to −∞ if it holds for no C.

Clearly, if the order is positive, then f (z) → 0 for z → ∞, uniformly in any verticalstrip. Conversely, if f is bounded in every vertical strip for z � 0, then the order of f at∞ is non-negative.

(4.2) Note. The order defined in (4.1) is in some sense analogous to the usual order at afinite point: Let f be a numeric function defined in an open neighborhood of a point u of C.Consider real numbers C such that

f (z) = O(|z − u|C) for z → u.

In other words, assume for some M and R that |f (z)| ≤ M|z − u|C for 0 < |z − u| < R.Clearly, ifC exists, then f (z) has finite values in a pointed neighborhood of u. The supremumof the possible numbers C is the order ordu f . Assume that f is holomorphic near u. Thenthe order is −∞ if u is an essential singularity of f . If f is meromorphic, then the order isthe usual order and, in particular, the order is an integer if f is not the zero function.

(4.3) Lemma. The order at ∞ of functions on H has the following properties:

(1) ordH∞(eiCz) = C.(2) ordH∞(f + g) ≥ inf{ordH∞ f , ordH∞ g}.(3) ordH∞(fg) ≥ ordH∞ f + ordH∞ g, with equality if ordH∞(1/f ) = − ordH∞(f ).(4) If z �→ αz is any Möbius transformation and k is a real number, then ordH∞ |αz|k = 0.(5) ordH∞ f (rz + b) = r ordH∞ f (z) for real numbers r > 0 and b.

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Proof. In (4), set f (z) := |αz|k . There are two cases to consider: α∞ �= ∞ and αz = ∞. Inthe first case, αz has a finite limit as z → ∞. In particular, (4.1.1) holds for C = 0 uniformlyon all of H. Thus the order is non-negative, and obviously, it is not positive. Hence the orderis equal to 0. In the second case, αz = cz + d with complex numbers c �= 0 and d. Clearly,in any vertical strip, we have |z| = O(z), and hence f (z) = O(|z|k). Hence (4.1.1) holdsfor any C < 0. Thus the order is non-negative, and obviously, it is not positive. Hence theorder is equal to 0.

The remaining assertions are easy consequences of the definition.

(4.4) Definition. Let f be a numeric function defined on a finite disk D. Let u be a point onthe boundary of D. Choose a Möbius transformation α mapping (H,∞) onto (D, u). Thenf will be said to be exponentially bounded at the point u, if the conjugate function f α on His exponentially bounded at ∞.

A second choice of α would differ from the first by a Möbius transformation of the formz �→ rz + b. Hence it follows from the result (4.3)(5) that the definition of exponentiallyboundedness is independent of the choice of α.

(4.5) Definition. In the setup of (4.4), from the same result (4.3)(5) quoted at the end, itfollows that we cannot define the order of f at u simply as the order of f α at ∞.

However, if a discrete subgroup of SL(D) is given, and u is a -parabolic point, we cannormalize the order and obtain the -order of f at u as follows: The conjugate group α isa discrete subgroup of SL(H), and the conjugate of the canonical generator γu is canonicalgenerator of α∞. Moreover, the latter canonical generator defines a Möbius transformationof the form z �→ z+ h, where h > 0. Set,

ord u f := h

2πordH

∞(f α). (4.5.1)

It is a consequence of the proof of the following lemma, that the -order at u is well defined.Obviously, it is different from −∞ if and only if f is exponentially bounded at u.

If u is a point of D, we define the -order of f at u by the formula,

ord u f := 1

|P u| ordu f, (4.5.2)

where the order on the right hand side is usual order at u of the function f , see (4.2). Notethat the order of the group P u is equal to the order of the canonical generator z �→ γu(z).If f is meromorphic, then the order on the right side of (4.5.2) is the usual order of f . Inparticular, then it is an integer (unless f = 0), and the -order of f at a point u of D is arational number. On the contrary, the -order at a cusp can be an arbitrary real number.

(4.6) Lemma. Consider a finite disk D, a discrete subgroup of SL(D) and a point u ofD ∪ ∂ D. Let z �→ αz be a Möbius transformation mapping a finite disk D ′ onto D. Letu′ := α−1u be the conjugate point and ′ := α be the conjugate subgroup. Consider afunction f on D and, for a given real number k, a weight-k conjugate f ′ = f α of f ,

f ′(z′) = ε

J (α, z′)kf (αz′). (4.6.1)

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[Autm] 19

Then,ord

′u′ f ′ = ord u f. (4.6.2)

Proof. Assume first that u belongs to D. The Möbius transformation z �→ αz is an analyticisomorphism. Hence, on the right side of (4.6.1), the order of f ′(αz′) at u′ is equal to theorder of f at u. Moreover, the fraction is holomorphic and everywhere non-zero on D′. Inparticular, its order at u′ is equal to 0. Hence, the order of f ′ at u′ is equal to the order of fat u. Moreover, the stabilizer groups ′

u′ and u are conjugate, and hence of the same order.Therefore, it follows from the definition (4.4) that (4.6.2) holds.

Assume next that u is -parabolic. Choose a Möbius transformation z �→ βz mapping(H,∞) onto (D, u). Then the conjugate subgroup � = α is a subgroup of SL(H), and theconjugate δ of the canonical generator of u defines a Möbius transformation z �→ z+hwithh > 0. By definition, the right hand side of (4.6.2) is equal to (h/2π) ordH∞ f (βz).

Choose and define similarly β ′, �′, δ′, and h′. By (4.6.1),

f ′(β ′z) = ε

J (α, β ′z)kf (αβ ′z). (4.6.3)

Now, the two matrices αβ ′ and β differ by the matrix σ := β−1αβ ′. The latter matrix belongsto SL(H), and it fixes the point ∞. Hence, σz = rz + b. Therefore, on the right side of(4.6.3) we have f (αβ ′z) = fβ(rz + b). Moreover, in the fraction, J (α, β ′z) is a Möbiustransformation or a constant function. Therefore, by Lemma (4.4), (4) and (5),

ordH∞(f

′β ′) = r ordH∞(fβ). (4.6.4)

Clearly, for the two subgroups � and �′ we have that �′ = σ−1�σ . Hence their canonicalgenerators δ′ and δ are conjugate: δ′ = σ−1δσ . Therefore, h′ = h/r . Now (4.6.2) followsfrom (4.6.4).

(4.7) Lemma. Let� be a subgroup of finite index in . Then, for any meromorphic functionf on D and any point u of D ∪ ∂ D,

ord�u f = |P u:P�u| ord u f.

Proof. Set d := |P u:P�u|. If u is a point of D, the assertion follows from the definitionand Lagrange: |P u| = d|P�u|. Assume that u is -parabolic. Then P u is an infinitecyclic subgroup. Therefore, if γu defines the canonical generator of P u, then γ du definesthe canonical generator of P�u. Hence, in the notation of Definition (4.5), if h is the hcorresponding to , then the h corresponding to � is equal to dh. Thus the assertion holdsfor a -parabolic point.

(4.8) Setup. Fix a finite disk D and a discrete subgroup of SL(D). Let j be a factor ofreal weight k on .

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A numeric function f defined on D is called a j -automorphic form, or a ( , j)-automor-phic form, if the following conditions hold:

(1) f is j -invariant: f (γ z) = j (γ, z)f (z) for γ ∈ ,(2) f is meromorphic in D,(3) f is exponentially bounded at all -parabolic points of D.

If the group is homogeneous, we will often assume that the factor j is homogeneous, sinceonly the zero function is j -invariant if j (−1, z) = −1, see (1.11)(4).

Note that to verify condition (3) for a given j -invariant function f , it suffices to verifythe condition for one point in each -parabolic orbit. Indeed, assume that f is exponentiallybounded at some point u on the boundary of D. Assume that u = γ u′ for a matrix γ in . It follows immediately from the definition in (4.4) that the function f γ is exponentiallybounded at the point u′. Since f is j -invariant, f (γ z) = j (γ, z)f (z). Therefore, it followsfrom Lemma (4.3)(4) that f is exponentially bounded at u′.

The condition (3) for a given meromorphic functionf is usually expressed by saying thatfis meromorphic at the cusps, and the automorphic forms defined above are sometimes said tobe meromorphic automorphic forms. An automorphic form f is said to be an integral form, iff is holomorphic in D and the -orders of f at all cusps are non-negative. If, in addition, the -orders at all cusps are positive, then f is called a cusp form. The spaces of j -automorphicforms that are meromorphic, or integral, or cusp forms, are denoted respectively,

M( , j), G( , j), S( , j).Clearly, the spaces are vector spaces over C.

If k is an integer, then J (γ, z)k is a factor of weight k on . More generally, then the factorsof weight k on are exactly the function j (γ, z) = χ(γ )J (γ, z)k for a unitary characterχ : → C∗. Accordingly, the corresponding spaces are denoted Mk( , χ), Gk( , χ), andSk( , χ). Omission of χ indicates the trivial character χ = 1. The forms of Mk( ) arecalled -automorphic forms of weight k.

When the weight k is equal to 0, a j -automorphic form is also called a j -automorphic func-tion. In particular, a -automorphic function is a -invariant function satisfying conditions(2) and (3). The space of -automorphic functions is denoted M( ).

(4.9) Observation. (1) The product j1j2 of factors of weights k1 and k2 on is a factorof weight k1 + k2. It follows that the product f1f2 of forms fi ∈ M( , ji) is a form inM( , j1j2). Similarly, the quotient f1/f2, when f2 �= 0, is a form in M( , j1/j2).

It follows in particular that the -automorphic functions form a field M( ). Moreover,if M( , j) �= (0), then M( , j) is a one-dimensional vector space over M( ). In otherwords, if f is a non-zero form in M( , j), then the map ϕ �→ ϕf is an isomorphism,

M( ) ∼−→M( , j).

(2) Let � be a subgroup of finite index in . Obviously, a ( , j)-invariant function is(�, j)-invariant. The two groups � and have the same set of parabolic points. Hence thecondition (4.8)(3) for a function f holds for if and only if it holds for �. Consequently,

M( , j) ⊆ M(�, j).

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[Autm] 21

Similarly, it follows from Lemma (4.7) that G( , j) ⊆ G(�, j) and S( , j) ⊆ S(�, j).(3) Consider a finite disk D and a Möbius transformation α : D′ → D defined by a matrix

α ∈ SL2(C). Then the conjugate factor jα is a factor of weight k on the conjugate group α = α−1 α. Choose a determination of J (α, z′)k on D′. Then for every function f on D,the weight-k conjugate function f α is a function onD′. Clearly, conjugationf �→ f α definesan isomorphismM( , j) ∼−→M( α, jα), and under conjugation, integral forms correspondto integral forms and cusp forms correspond to cusp forms.

(4.10) Lemma. Let f be a ( , j)-automorphic form on D. If u and v are -equivalent pointsof D ∪ ∂ D, then

ord u f = ord v f. (4.10.1)

Proof. The equation follows from Lemma (4.6). Indeed, let v = γ u for some matrix γ in . Then (4.10.1) holds if the function f on the left hand side is replaced by any weight-kconjugate f γ of f . However, the function f ·j γ is weight-k conjugate, and it is equal tof .

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[Autm] 23

5. Fourier expansions.

(5.1). Keep the setup of (3.1). Assume for the moment that D is the upper half plane Hand that the point ∞ on the boundary of H is -parabolic. Then the Möbius transformationassociated to the canonical generator γ∞ is of the form,

γ∞z = z+ h, for some h > 0.

The sign, ω∞(j), of j at infinity is of modulus 1, and the parameter κ = κ∞(j) is defined bethe conditions,

ω∞(j) = e2πiκ , 0 ≤ κ < 1.

Since J (γ∞, z) = ±1, it follows that j (γ∞, z) is constant,

j (γ∞, z) = e2πiκ .

Now, let f be a non-zero j -automorphic form on H. Then, in particular, f (γ∞z) =j (γ∞, z)f (z), that is,

f (z+ h) = e2πiκf (z). (5.1.1)

Let F(z) be the function in H defined by the equation,

F(z) := e−2πiκz/hf (z). (5.1.2)

It follows from (5.1.1) that F is periodic:

F(z+ h) = F(z).

In addition, the function F(z) is meromorphic because f (z) is. Consequently, there exists aunique meromorphic function g(q) on the pointed unit disk: 0 < |q| < 1 such that

F(z) = g(e2πiz/h). (5.1.3)

The function F(z) is exponentially bounded because f (z) is. In particular, in a strip of widthh, it follows that F(z) has no poles for z > R. Hence, being periodic, it follows that F(z)is holomorphic in the domain z > R. Therefore, the function g is holomorphic in a pointeddisk: 0 < |q| < ε, and the possible singularity of g at 0 is isolated. It follows that g has aconvergent Laurent expansion,

g(q) =∑

−∞<n<∞anq

n for 0 < |q| < ε. (5.1.4)

The possible singularity 0 of g is at most a pole, that is, the Laurent expansion (5.1.4) hasonly a finite number of negative terms. Indeed, since F(z) is exponentially bounded, there isa real number C, say of the form −2πN/h for some integerN , so that F(z)/eiCz is bounded

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for z > R. It follows that qNg(q) is bounded near 0. Hence, in the Laurent expansion,an = 0 for n < N . Thus the function g is meromorphic in the whole unit disk: |q| < 1, andthe order, ord0 g, of g at 0 is the least integer n such that an �= 0.

LetN be the order of g at 0. Then g(q) = qN g(q), where g is holomorphic and non-zeroat 0. Hence F(z) = e2πiNz/hg(e2πiz/h). Obviously, then ordH∞ F(z) = 2πN/h. Combinedwith the definition of F , it follows that

ord ∞ f = κ +N.

To summarize, we have proved the following: There is a Laurent series,

g(q) =∑n≥N

anqn, (5.1.5)

convergent in a pointed disk 0 < |q| < ε, so that for z � 0 (more precisely, for z > R

where R = (− log ε)h/2π ) we have the following expansion of f ,

f (z) = e2πiκz/h∑n≥N

ane2πinz/h (5.1.6)

Clearly, the -order of f at ∞ is related to the order of g at 0 by the formula,

ord ∞ f = κ + ord0 g (5.1.7)

(5.2) Definition. The series (5.1.6), defined in the setup of (5.1), is called the Fourier expan-sion of the form f , and the coefficients an are called the Fourier coefficients. The functionq = q(z) = e2πiz/h is called the local parameter at ∞. As a function onH it has, for any realnumber l the obvious determination q l = e2πilz/h. Accordingly, we may write the Fourierexpansion in the form,

qκ∑n≥N

anqn. (5.2.1)

Note that (5.2.1) does not make sense as a function of q for 0 < |q| < ε unless κ is an integer.

Now assume again thatD is an arbitrary finite disk, and let f be a non-zero j -automorphicform on D. Let u be a -parabolic point. Consider the sign ωu = ωu(j) and the parameterκu = κu(j). Choose a Möbius transformation z �→ βz mapping (H,∞) onto (D, u), andchoose a determination of J (β, z)k on H. Then the weight-k conjugate,

f β(z) = 1

J (β, z)kf (βz),

is a jβ-automorphic form on H. Hence the preceding discussion applies to f β . Underconjugation, the sign and the parameter are unchanged by (3.11), and the order is unchangedby (4.8). The series (5.2.1) for f β , or the series (5.1.6), is called the Fourier series of f atthe point u. It follows from (5.1.7) that,

ord u f = κu + ord0 g.

In particular, the -order ord u f of a j -automorphic form f is congruent to the parameter κumodulo Z.

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[Autm] 25

(5.3). Assume for the moment that D is the unit disk E, and consider the point 0 in E. TheMöbius transformation associated to the canonical generator γ0 is of the form,

γ0z = e2πi/ez, where e = |P 0|.

[Warning: there are two e’s: one for Euler, and one for elliptic.] The sign, ω0(j), of j at 0is an e’th root of unity. The parameter κ = κ0(j) is defined be the conditions,

ω = e2πiκ , 0 ≤ κ < 1,

and hence κ = a/e for some integer a with 0 ≤ a < e. The matrix γ0 is a diagonal matrix,and hence J (γ0, z) is constant. It follows that j (γ0, z) is constant,

j (γ0, z) = e2πiκ .

Now, let f be a non-zero j -automorphic form on E. Then, in particular, f (γ0z) =j (γ0, z)f (z), that is,

f (e2πi/ez) = e2πiκf (z). (5.3.1)

Let F(z) be the function in E defined by the equation,

F(z) := z−eκf (z); (5.3.2)

note that eκ is an integer, since κ = a/e. It follows from (5.3.1) that F(z) satisfies theequation:

F(e2πi/ez) = F(z).

In addition, the function F(z) is meromorphic because f (z) is. Consequently, there existsa unique meromorphic function g(w) on the pointed unit disk 0 < |w| < 1 such that, forz �= 0,

F(z) = g(ze). (5.3.3)

The possible singularity for g at 0 is a pole. Indeed, since F(z) is meromorphic at 0, thereis an integer, say of the form eN , so that zeNF (z) is bounded near 0. Hence, it follows from(5.3.2) that wNg(w) is bounded near 0. Therefore, the function g has a convergent Laurentexpansion in a small pointed disk,

g(w) =∑n≥N

anwn for 0 < |w| < ε, (5.3.4)

Assume that N is the order of g at 0, i.e., aN �= 0. Then it follows from (5.3.3) F(z) is oforder eN at 0, and then from (5.3.1) that f is of order eκ + eN . Hence, for the -order, weobtain the equation,

ord 0 f = κ +N.

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To summarize, we have proved the following: The Laurent series for f near 0 is of the formf (z) = zeκg(ze), that is,

f (z) = zeκ∑n≥N

anzen, (5.3.5)

where e := |P 0|. The -order of f at 0 is related to the order of g at 0 by the formula,

ord 0 f = κ + ord0 g (5.3.6)

(5.4) Definition. Assume again that D is an arbitrary finite disk, and let f be a non-zeroj -automorphic form on D. Let u be a point of D. Consider the sign ωu = ωu(j) and theparameter κu = κu(j). Choose a Möbius transformation z �→ βzmapping the unit disk (E, 0)onto (D, u), and consider as in (5.2) a weight-k conjugate f β . Then f β is a jβ-automorphicform on E. Hence the preceding discussion applies to f β . The Laurent series (5.3.5) forf β , or the expansion (5.3.4), is called the normalized Laurent series for f at the point u. Itfollows from (5.3.5) that,

ord u f = κu + ord0 g.

In particular, the -order ord u f of a j -automorphic form f is congruent to the parameter κumodulo Z.

(5.5) Note. Let f be a ( , j)-automorphic form on D, and let u be a point of D∪ ∂ D. Thenthe Fourier expansion of f at u defined in (5.2) when u is -parabolic and the normalizedLaurent expansion defined in (5.4) when u belongs to D depend on the choice of conjugationβ. (In addition, they depend on the choice of a determination of J (β, z)k, but the latterambiguity is only up to a complex sign.) To normalize, we may assume that det β = 1.

Consider the case when u is -parabolic. Then a second choice β ′ is of the form βσ ,where σz = rz + b (r > 0). Clearly then, a weight-k conjugate f β

′can be obtained as a

weight-k conjugate (f β)σ . Hence, to simplify, we may assume that D = H and that β = 1.Then β ′ = σ and the canonical generator of σ has h′ = h/r . We want to compare theFourier expansion (5.1.6) of f with the Fourier expansion of the weight-k conjugate f σ . Thematrix σ is a diagonal matrix. Hence J (σ, z) is constant, and equal to ±1/

√r . Hence, up

to a complex sign, the determination J (σ, z)k is constant and equal to r−k/2. Therefore, aweight-k conjugate f σ is given by

f σ (z) = rk/2f (rz+ b).

Accordingly, we obtain the Fourier expansion of f σ essentially by substitution of rz+ b forz in (5.1.6). If we let λ := rk/2e2πiκb/h and ε := e2πib/h, then the result is the following:

f σ (z) = λe2πiκz/h′ ∑n≥N

(εnan)e2πinz/h.

In particular, the Fourier coefficients of the new series are given by

a′n = λεnan.

Here |λ| = rk/2 and |ε| = 1.

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(5.6) Example. The Eisenstein series Ek(z), for an integer k ≥ 4, and the normalized series,

Gk(z) := 1

2

∑(m,n)=1

1

(mz + n)k,

were considered in Example (2.1). They are (1)-invariant functions of weight k. Clearly,they are holomorphic. The Fourier expansions of the functions are determined in (App.2.5).The series vanish when k is odd. For k ≥ 4 even, the expansion is the following, for q = e2πiz,

Gk(z) = 1 + −2k

Bk

∑n≥1

σk−1(n)qn, (5.6.1)

where Bk is the k’th Bernoulli number. In particular, the order at ∞ of the functions are equalto 0. As ∞ represents the only cusp of (1), it follows that the functions Ek(z) and Gk(z)are integral automorphic forms.

For k = 4 and k = 6, the expansions are the following,

G4(z) = 1 + 240∑r≥1

σ3(r)qr ,

G6(z) = 1 − 504∑r≥1

σ5(r)qr .

The functionsG4(z)3 andG6(z)

2 are integral automorphic forms of weight 12. Clearly, intheir Fourier expansion, the constant term is equal to 1. Hence, for the difference, G4(z)

3 −G6(z)

2, there is no constant term in the expansion, that is, the difference is a cusp form ofweight 12 for (1). It is easy to see that the coefficient to q is equal to 1728 = 123. So thedifference has an expansion of the form,

G4(z)3 −G6(z)

2 = 123q + · · · . (5.6.2)

(5.7) Example. Dedekind’s η-function,

η(z) = e2πiz/24∏n≥1

(1 − e2πinz),

was considered in Example (2.3). It is holomorphic inH and everywhere non-zero. The factorjη is of weight 1

2 on (1) and, by its construction, the function η is jη-invariant. Obviously,the order, ordH∞ η, at ∞ is equal to 2π/24. As ∞ represents the only cusp for (1), it followsthat η is a jη-automorphic form. The canonical generator at ∞ is the matrix t , and tz = z+1.Hence h = 1, and the (1)-order of η at ∞ is equal to 1/24. As the order is positive, itfollows that η is a cusp form. At the points of H, the function η is non-zero, that is, the orderis equal to 0.

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The Fourier expansion of η is obtained by developing the product∏n≥1(1 − qn) into a

power series. With its first terms it becomes,

η(z) = q1/24(1 − q − q2 + q5 + q7 − · · · ). (5.7.1)

In fact, the coefficients in the expansion are given by Euler’s well known pentagonal numberformula, ∏

n≥1

(1 − qn) =∑m

(−1)mq(3m2−m)/2.

It follows from the expansion of η that the discriminant�(z) defined in Example (2.3) asthe power �(z) = η(z)24 has a Fourier expansion of the form,

�(z) = q − 24q2 + · · · . (5.7.2)

We prove later that the difference (5.6.2) is in fact equal to 123 times the discriminant (5.7.2).

(5.8) Example. The θ -function,

θ(z) =∑n

eπin2z,

was considered in Example (2.6). It is holomorphic in H. The factor jθ is of weight 12 on

the group θ , and by its construction, the function θ is jθ -invariant. There are two cusps for θ , represented by the points ∞ and −1. Clearly, the θ -function converges uniformly to 1as z → ∞. In particular, the θ -function is exponentially bounded at ∞ and the order at∞ is equal to 0. Essentially, the sum defining θ is the Fourier expansion at ∞. In fact, at∞ the canonical generator of θ is t2, and hence h = 2. Thus the local parameter at ∞ isq = e2πiz/2 = eπiz, and the Fourier expansion at ∞ is the series,

1 + 2q + 2q2 + 2q4 + 2q9 + · · · . (5.8.1)

For the second cusp −1, take the conjugation by the matrix u. Then u∞ = −1. A weight- 12

conjugate θu is then given by the equation,

θu(z) = ε√J (u, z)

θ(uz) = ε√zθ(−1 − 1/z).

for any complex sign ε. In particular, the function√i/z θ(−1−1/z) is a conjugate. It follows

from (2.7.3) that the latter conjugate has the expansion,

θu(z) =∑n odd

eπin2z/4. (5.8.2)

In particular, the conjugate is exponentially bounded at ∞, and hence θ is exponentiallybounded at −1. Therefore, θ is a jθ -automorphic form. To get the Fourier expansion of θ at−1, note that under conjugation by u, the canonical generator of θ at −1 corresponds to t .

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[Autm] 29

Hence h = 1, and the local parameter at ∞ is q = e2πiz. Now, when n is odd, we have thatn2 ≡ 1 modulo 8. The n’th term in (5.8.2) is equal to e2πiz(n2−1)/8e2πiz/8. Thus (5.8.2) isthe Fourier expansion of θ at −1, and it takes the form,

θu(z) = e2πiz/8∑n odd

e2πiz(n2−1)/8 = q1/8(2 + 2q + 2q3 + 2q6 + 2q10 + · · · ).

In particular,ord θ−1 θ = 1/8.

As the orders at the two cusps are nonnegative, θ is an integral form.

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[Autm] 31

6. The Main Theorems.

(6.1). Keep the setup of (3.1). In other words, D is a finite disk, is a discrete subgroup ofSL(D), and j is a factor of real weight k for . In this section, we assume throughout that is a Fuchsian group of the first kind. In addition, we assume that the factor j is homogeneous,that is, if −1 ∈ , then j (−1, z) = 1.

The group acts on D and on the closure D ∪ ∂ D. Since is of the first kind, the orbitspaceX = D/ is a compact surface. Moreover, there exists a finite fundamental domain Ffor . Accordingly, there are two fundamental invariants associated with . The first is thenonnegative integer g = g( ) defined as the genus of X. The second is the positive rationalnumber μ = μ( ) defined as the area of F divided by 2π . The two invariants are related bythe formula, see (Discr.5.5),

μ( ) = 2g( )− 2 +∑

u mod

(1 − 1

eu

), (6.1.1)

where eu = |P u|. In the sum (6.1.1), and in the sums below, the summation is over the orbitsin X, or equivalently, over one point u in each -orbit in D ∪ ∂ D. For a -parabolic pointu, the fraction 1/eu is interpreted as zero. So, in the sum (6.1.1), each of the finitely manycusps contributes with the term 1, and for the orbits in D, only the finitely many -ellipticorbits contribute with non-zero terms.

In this section we state the main theorems on automorphic forms. The proofs are postponedto a separate section.

(6.2) Theorem A. The field M( ) of -automorphic functions on D is finitely generated (asa field) over C, and of transcendence degree 1.

The quotient X = X( ) is a compact Riemann surface, and by construction, M( ) isthe field of meromorphic functions on X. The assertion of Theorem A is a well knownconsequence.

(6.3) Theorem B. The space M( , j) of j -automorphic forms on D is non-zero. Moreover,if f is any non-zero j -automorphic form, then

∑u mod

ord u f = kμ( )

2. (6.3.1)

The existence of a non-zero j -automorphic form will follow from the construction ofPoincaré Series. The proof of (6.3.1) will be given later in this chapter.

(6.4) Corollary. The space G( , j) of integral j -automorphic forms is of finite dimensionover C. If k < 0, then the dimension is equal to zero. If k > 0 then the dimension is at mostequal to kμ/2 + 1. If k = 0, in which case j (γ, z) = χ(γ ) with a unitary character χ on ,then the dimension is zero if χ �= 1. Finally, the functions of G( ) are exactly the constantfunctions.

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Proof. Consider a nonzero function f in G( , j). Then the orders ord u f are non-negative,and the sum in (6.3.1) is equal to kμ/2. Clearly, therefore, no such f can exist if k < 0.

Assume k ≥ 0. Then ord u f ≤ kμ/2. Let N be the largest integer less than or equal tokμ/2. To prove that the dimension is finite we may assume that the disk is the unit disk E.Since the order of f at zero is nonnegative, the normalized Laurent series around 0 is of theform,

f (z) = zeκ∑n≥0

anwn for w = ze, (6.4.1)

where κ = κ0(j) and e = e0. The -order of f at zero is equal to κ + n where n is leastindex such that an �= 0; on the other side, the -order is at most equal to kμ/2. It followsthat one of the N + 1 coefficients an for 0 ≤ n ≤ kμ/2 is nonzero. The N + 1 coefficientsmay be viewed as a C-linear map G( , j) → CN+1, and we have just seen that the map isinjective. Hence the dimension of G( , j) is finite and at most equal to N + 1.

Assume that k = 0. Let f be a function in G( , χ). Then f is holomorphic in D, anddefines a continuous function on D ∪ ∂ D. Moreover, since χ is unitary, it follows from theequation f (γ z) = χ(γ )f (z) that the function |f | is -invariant. As the quotient D/ iscompact, it follows that |f | attains its maximum value at a point u of D ∪ ∂ D. It followsfrom the maximum principle that f is constant. Indeed, if u is in D, the maximum principleapplies directly. If u is in ∂ D, we may, after conjugation, assume that D = H and u = ∞.If κu = 0 then, in a neighborhood of u, f = g(q) where q = e2πiz/h and g is holomorphicat the origin; hence the maximum principle applies to g. If κu > 0, then f vanishes at u; bythe choice of u, therefore f = 0.

Hence every function in G( , χ) is constant. Clearly, if χ �= 1, then no non-zero constantis χ -invariant.

Thus all assertions of the Corollary have been proved.

(6.5) Note. Assume that k > 0. It follows from (6.1.1) that the number kμ/2 is at most equalto the following number,

k(g − 1)+ k

2#(parabolic or elliptic -orbits). (6.5.1)

Hence, ifN is the integer part of the number (6.5.1), then the dimension of G( , j) is at mostN + 1. Clearly, the argument given in the proof of (6.4) applies equally well to a -parabolicpoint u, replacing the Laurent expansion by the Fourier expansion. In particular, if for twofunctions f and g in G( , j) it is known that their first N + 1 Fourier coefficients are equal,then the two functions are equal.

(6.6) Special case. Assume that the disk is the upper half plane H and that is a subgroupof finite index in the modular group (1) = SL2(Z). Denote by d the homogeneous index,d = | (1) : |. Then, for any non-zero j -automorphic form on H,

∑u mod

ord u f = kd

12. (6.6.1)

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[Autm] 33

Moreover, if k > 0, then the dimension of G( , j) is at most equal to [kd/12] + 1.

Proof. In the general setup of (6.1), let� be a subgroup of finite index in . Set d := |P :P�|.Then, by Proposition (Discr.3.12), a fundamental domainG for� can be obtained as the unionof d transforms of the fundamental domain F for . Hence the area of G is d times the areaof F . Therefore, μ(�) = d μ( ).

A fundamental domain for the modular group (1) was described in (Mdlar.4.1). Clearly,its area is equal to 2π/6. Consequently, μ( (1)) = 1

6 . Therefore, the assertion is a specialcase Theorem B and its Corollary.

(6.7) Theorem C. The number δ = δ( , j), defined by the following expression, is an integer:

δ := 1 − g + kμ

2−

∑u mod

κu(j), (6.7.1)

where κu(j) is the parameter of j at u. If δ < 1 − g, then G( , j) = (0). Assume thatδ ≥ 1 − g. Then the following inequalities hold:

δ ≤ dim G( , j) ≤ δ + g.

Moreover, if δ ≥ g, thendim G( , j) = δ. (6.7.2)

The number δ is an integer. Indeed, let f be a non-zero j -automorphic form (TheoremA). By Theorem B, the number kμ/2 is the sum of the -orders of f . As observed in Section5, the -order ord u f is congruent modulo Z to the parameter κu(j). Thus the differenceord u f − κu(j) is an integer. Hence δ is an integer.

If f is an integral non-zero j -automorphic form, then ord u f ≥ 0. As the order iscongruent to κu and 0 ≤ κu < 1, it follows that ord u f ≥ κu(j). Hence δ ≥ 1 − g.Therefore, the assertion that G( , j) = (0) when δ < 1 − g is a consequence of Theorem B.

The remaining assertions of Theorem C are consequences of Riemann’s part of the Rie-mann–Roch Theorem. Note that the assertions give full information on the dimension ofG( , j) when g = 0. If g > 0, the Theorem gives only an estimate of the dimension when1 − g ≤ δ ≤ g − 1.

(6.8) Note. The following alternative expressions for δ are easily obtained from the formula(6.1.1):

δ = g − 1 + (k2 − 1

)μ+

∑(1 − 1

eu− κu

)

= (k − 1)(g − 1)+∑(

k2

(1 − 1

eu

) − κu

).

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(6.9) Corollary. The dimension of G( , j) is equal to δ and at least equal to g under any ofthe following conditions: (1) k > 2, or (2) k = 2 and there are -parabolic points, or (3)k = 2 and there are -elliptic points u at which κu(j) < 1 − 1/eu.

Proof. Consider the first expression for δ in (6.8). In the sum, the term corresponding toa -parabolic point u is equal to 1 − κu, and hence positive. The term corresponding to a -elliptic point is non-negative by (3.3), and positive if κu < 1 − 1/eu. It follows easily theδ > g − 1 under any of the conditions given for the first assertion. Therefore, since δ is aninteger, the first assertion of the Corollary follows from Theorem C.

(6.10) Definition. A -parabolic point u is said to represent a j -regular cusp, if the signωu(j) is equal to 1, or, equivalently, if κu(j) = 0. Obviously, j -regularity is a property ofthe cusp represented by u.

It follows from (3.13) that a cusp is J -regular if and only if it is regular. Moreover, if k isan even integer, then all cusps are J k-regular. If k is an odd integer, then a cusp is J k-regularif and only if it is a regular cusp for .

Denote by δ′ = δ′( , j) the following number,

δ′ = δ − #(j -regular cusps).

(6.11) Theorem D. Consider the space S( , j) of j -automorphic cusp forms on D. Ifδ′ < 1 − g, then S( , j) = (0). Assume that δ ′ ≥ 1 − g. Then the following inequalitieshold:

δ′ ≤ dimS( , j) ≤ δ′ + g.

Moreover, if δ′ ≥ g, thendimS( , j) = δ′. (6.11.1)

(6.12) Corollary. The dimension of S( , j) is equal to δ ′ and at least equal to g under anyof the following conditions: (1) k > 2, or (2) k = 2 and there are -parabolic points thatare not j -regular, or (3) k = 2 and there are -elliptic points u at which κu(j) < 1 − 1/eu.

Proof. Consider the first expression for δ in (6.8). Clearly, in the sum, each regular cusp ucontributes with the term 1. Therefore, if we omit from the sum the terms corresponding tothe regular cusps, the resulting expression is equal to δ ′. The remaining part of the proof isnow identical to the proof of Corollary (6.9).

(6.13) Note. An integral j -automorphic form f has non-negative -order at every point u ofD ∪ ∂ D. In particular, f has a well defined value at points u of D. For a point u ∈ ∂ D,choose a conjugation α : (H,∞) → (D, u). Then the weight-k conjugate function f α has avalue at ∞ which, by an abuse of language, will be referred to as the value of f at u. Thevalue is zero if and only if the order of f at u is positive. The order is at least equal to theparameter κu. In particular, the value at a point u representing a j -irregular cusp is necessarilyzero. It follows that f is a cusp form if and only if the value is equal to zero at every j -regular

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[Autm] 35

cusp. In other words, if Zreg is a finite set of -parabolic points representing the subset ofj -regular cusps, then the evaluation map,

G( , j) → CZreg, (6.10.1)

has as kernel the space S( , j) of cusp forms. In particular, the codimension of S( , j) inG( , j) is at most equal to the number of j -regular cusps, and equality holds if and only ifthe evaluation map is surjective. Clearly, the evaluation map is surjective, if and only if, forevery j -regular cusp u for , there exists an integral j -automorphic form that has a non-zerovalue at u and vanishes at all other cusps.

Since δ − δ′ is equal to the number of j -regular cusps, it follows that if both equalities(6.7.2) and (6.11.1) hold, then the evaluation map is surjective. In particular, therefore theevaluation map is surjective if δ ′ ≥ g, and especially, it is always surjective when k > 2. Itshould be emphasized that the evaluation map is not surjective in general.

(6.14). Consider in particular the factor j = J k . Thus k is assumed to be an integer; ifk is odd, the group is assumed to be inhomogeneous. The parameters at cusps u wereconsidered in (6.10): If k is even, then all cusps are J k-regular (and κu = 0). If k is odd, thenthe J k-regular cusps are exactly the regular cusps for . At the irregular cusps, the parameteris equal to 1

2 .Consider the parameter at a point u of D. It follows from (3.13) that κu(J k) is equal to{

k2 (1 − 1/eu)

}, where {x} denotes the fractional part, {x} := x − [x].

An expression for δk = δ( , J k) is obtained from the second expression for δ in (6.8). Ifk is even,

δk = (k − 1)(g − 1)+∑[

k2

(1 − 1

eu

)]. (6.14.1)

Note that each cusp of contributes with the integer k/2 to the sum. If k is odd,

δk = (k − 1)(g − 1)+∑[

k2

(1 − 1

eu

)] + 12 #(regular cusps). (6.14.2)

Note that each cusp contributes with the integer[k2

] = k2 − 1

2 to the sum.The numbers for k = 1 and k = 2 are the following:

δ1 = 1

2#(regular cusps), δ2 = g − 1 + #(cusps). (6.14.3)

The dimensions of Gk( ) and Sk( ) are determined by the preceding results for k ≤ 0and k ≥ 3. In addition, if there are cusps for , then the results imply that dim G2( ) = δ2.

(6.15) Theorem E. The dimension ofS2( ) is equal to g. In addition, if is inhomogeneous,then the codimension of S1( ) in G1( ) is equal to half the number of regular cusps.

The proof of Theorem E uses the Roch part of the Riemann–Roch Theorem.Note that the dimension of G2( ) is determined in all cases. If there are cusps, then

the dimension is equal to the number δ2 of (6.14.3). If there are no cusps, then of courseG2( ) = S2( ), and the dimension is equal to g.

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As noted in (6.13), the codimension ofSk( ) in Gk( ) is equal to the number of J k-regularcusps if k ≥ 3. It follows from Theorem E that the assertion does not hold for k = 2 if thereare cusps, and it does not hold for k = 1 if there are regular cusps.

Theorem E determines the codimension of S1( ) in G1( ), but no general method isknown to determine the actual dimension.

(6.16) Theorem F. Assume that the disk is the upper half plane H. Let f be a non-zeroj -automorphic cusp form. Then the function f (z)(z)k/2 is bounded in H. Consider theFourier expansion of f around a cusp u,

f α = e2πiκz/h∞∑n=0

ane2πinz/h (6.16.1)

where κ = ord u f is positive. Then, for the coefficients, we have the estimate,

an = O(nk/2). (6.16.2)

Proof. Note that the weight k is non-negative, since S( , j) �= (0). Consider the functionρ(z) = |f (z)|(z)k/2. As (γ z) = |J (γ, z)|−2(z) and f is j -invariant, it follows that ρ(z)is -invariant. Obviously, ρ is continuous in D. We claim that ρ(z) converges to zero whenz approaches a cusp u. Clearly, if α is a matrix of SL(H) such that α∞ = u, it suffices toprove that ρ(αz) → 0 uniformly as z → ∞. Clearly,

ρ(αz) = |f (z)|(αz)k/2 = |f (z)||J (α, z)|k (z)

k/2. (6.16.3)

The fraction on the right is the absolute value of a weight-k conjugate f α . It follows fromthe Fourier expansion (6.16.1) that f α = O(|e2πκz/h|). Thus ρ(αz) → 0 for z → ∞.

It follows that ρ extends to a continuous function on the quotient D/ . As the quotient iscompact, therefore ρ is bounded on H.

Assume that ρ(z) ≤ K for all z in H. Let g(q) be the function defined in the unit disk:|q| < 1 by f α(z) = e2πiκz/hg(e2πiz/h). Then g is holomorphic in the unit disk, since fin holomorphic in H. The Fourier coefficients an are the coefficients in the power seriesexpansion of g. Hence they can be obtained by integration,

an = 1

2πi

∫g(q)

qn

dq

q,

where the path integral can be taken over any circle around 0 in the unit disk. Equivalently,

an = 1

h

∫g(e2πiz/h)dz

e2πinz/h = 1

h

∫f α(z)dz

e2πi(n+κ)z/h ,

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[Autm] 37

where the path integral can be taken over any (euclidean) vertical line segment of length h inH. Integrate from iε to iε + h. On the path we have, by (6.16.3), the estimate,

|f α(z)| = ρ(αz)(z)−k/2 ≤ Kε−k/2,

and the equality |e2πi(n+κ)z/h| = e−2π(κ+n)ε/h . Therefore, we obtain the estimate for thecoefficient,

|an| ≤ Kε−k/2e2π(κ+n)ε/h.

Fix ε and apply the estimate with ε := ε/n for n ≥ 1. Since (κ + n)/n ≤ 2, it follows that

|an| ≤ Kε−k/2e4πε/h nk/2.

Therefore (6.16.2) holds.

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[Autm] 39

7. Examples II.

(7.1) Example. For the modular group (1) = SL2(Z) acting on H, there are two ellipticorbits represented by i of order 2 and ρ of order 3, and one cusp represented by ∞. Thegenus is g = 0; the area is π/3, and so μ = 1

6 , confirming that16 = 0 − 2 + (1 − 1

2 )+ (1 − 13 )+ 1.

Let k be a non-negative even integer, and consider the spaces Gk := G( (1), J k) and Sk :=S( (1), J k). Apply the Main Theorems to the factor J k . The number δk = δ( (1), J k) isgiven by (6.14.1):

δk = −(k − 1)+ [k2

(1 − 1

2 )] + [

k2

(1 − 1

3 )] + k

2 ={

1 + [k12

]when k �≡ 2 (mod 12),[

k12

]when k ≡ 2 (mod 12).

Note that δk ≥ 0 = g since k is non-negative. Hence it follows from Theorem C that

dim Gk = δk.

Clearly, if k ≥ 4, then δk − 1 ≥ 0 = g. Therefore, by Theorem D, if k ≥ 4, then

dim Sk = δk − 1.

Obviously, S0 = 0, and by Theorem E,

dimS2 = 0.

For small values of k, the dimensions are the following:

k 0 2 4 6 8 10 12 14 16 18 20 22 24 26 28dim Gk 1 0 1 1 1 1 2 1 2 2 2 2 3 2 3dimSk 0 0 0 0 0 0 1 0 1 1 1 1 2 1 2

The Eisenstein seriesEk(z) and modified seriesGk(z), for k ≥ 4, are functions inGk . Theyare not cusp forms, since Gk has the value 1 at ∞. Clearly, the functionGk(z) generates the1-dimensional space Gk for k = 4, 6, 8, 10, 14. In particular, comparing the values at ∞, weobtain the equations G8(z) = G4(z)

2, G10(z) = G4(z)G6(z), and G14(z) = G4(z)2G6(z).

The first weight k at which there is a non-trivial cusp form is k = 12. The discriminant�(z), defined by �(z) = η(z)24, is a cusp form of weight 12, with the Fourier expansion�(z) = q − 24q + · · · . Hence it generates the 1-dimensional space S12. On the other hand,the difference G3

4 −G26 is a cusp form; it is easily seen that the coefficient to q in its Fourier

expansion is equal to 1728. Therefore, since the space S12 is 1-dimensional, the followingequation is a consequence:

�(z) = 12−3(G 34 −G 2

6

). (7.1.1)

The Fourier expansion of �(z) is of the following form,

�(z) =∑n≥1

τ(n)qn, with τ(1) = 1. (7.1.2)

The function τ(n) is Ramanujan’s τ -function. From the equation �(z) = η(z)24, it followsimmediately that τ(n) ∈ Z. It is not hard to see directly the right hand side of (7.1.1) has aFourier expansion with integral coefficients.

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(7.2) Example. The Fourier expansions of the functions Gk(z) for k = 4, 6, 8, . . . aredetermined in (App.2.6): If we write �k := ∑

r≥1 σk(r)qk , then for even k ≥ 4,

G(z) = 1 + αk�k−1, (7.2.1)

where αk = −2k/Bk . In terms of the integers Ak of (App.1),

αk = (−1)k/22k+1(2k − 1)

Ak.

In particular, as A4 = 2 and A6 = 24, it follows that α4 = 24 · 3 · 5 = 240 and α6 =−23 · 32 · 7 = −504. The number αk is the coefficient of q in the Fourier expansion (7.2.1).In particular, from the equations G8 = G 2

4 , G10 = G4G6, and G14 = G8G6 observed in(7.1), it follows that α8 = 2α4 = 25 · 3 · 5, α10 = α4 + α6 = −23 · 3 · 11, and α14 = −23 · 3.Of course, the values of α8 and α10 confirm the values A8 = 24 · 17 and A10 = 28 · 31 in(App.1.4). Since A12 = 29 · 691, it follows that

α12 = 24 · 32 · 5 · 7 · 13

691. (7.2.2)

Relations among the series Gk(z) imply relations among their Fourier coefficients. Forinstance, from the relation G8 = G 2

4 , it follows that α8�7 = (α4�3)2 + 2α4�3. Hence,

σ7(r) = σ3(r)+ 23 · 3 · 5r−1∑t=1

σ3(t)σ3(r − t).

For the discriminant�(z), say given by (7.1.1), it follows that

(12)3�(z) = G 34 −G 2

6 = α34�

33 + 3α 2

4�23 + 3α4�3 − α 2

6�25 − 2α6�5.

As a consequence,

�(z) = 5�3 + 7�5

12+ 8000� 3

3 + 100� 23 − 147� 2

5 .

It is easily seen that the fraction has integer coefficients. Hence the coefficients τ(n) of�(z)are integers. It follows easily from the equation that τ(1) = 1, τ(2) = −24, and τ(3) = 252.It can be proved that the function τ(n) is multiplicative. Moreover, for a primep the followingequation holds:

τ(pt+1) = τ(p)τ(pt )− p11τ(pt−1).

Thus τ(n) is completely determined by its values on primes p. Since �(z) is a cusp form ofweight 12, Theorem F implies the estimate τ(n) = O(n12/2). In fact, Ramanujan’s conjecture,

|τ(p)| < 2p11/2 for all primes p,

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[Autm] 41

was proved by P. Deligne in 1974. Far more trivial is Ramanujan’s congruence,

τ(n) ≡ σ11(n) (mod 691).

It can be proved as follows. The form G12 − G 34 is a cusp form of weight 12, and hence

equal to a constant times �. The constant can be determined by comparing the coefficientsto q. It follows that G12 − G 3

4 = (α12 − 3α4)�. By (7.2.2) the number a := 691α12 is aninteger, and prime to 691. Hence, by multiplying by 691, we obtain the equations of formswith integer coefficients:

(a − 691 · 3 · α4)� = 691G12 − 691G 34 = a�11 − 691(G 3

4 − 1).

Modulo 691, it follows that aτ(n) ≡ aσ11(n). Hence Ramanujan’s congruence holds.

(7.3) Example. Consider again the modular group (1). Let f be a non-zero function inMk = Mk( (1)). By the Special Case (6.6), we have the equation

∑u ord (1)u f = k/12.

The (1)-order at ∞, which we will denote by ord∞ f , is an integer, since the parameter at∞ is equal to zero. The (1)-order at i is equal to 1

2 ordi f and the (1)-order at ρ is equalto 1

3 ordρ f . At points of H that are not (1)-equivalent to i or ρ, the (1)-order is simplythe usual order. Hence the equation may be written as follows:

∑u

ordu f + 12 ordi f + 1

3 ordρ f + ord∞ f = k12 , (7.3.1)

where the sum is over a system of representatives for the (1)-ordinary orbits. If f is anintegral form, then the orders are non-negative.

For example, consider a non-zero form f in S12. The right hand side of (7.3.1) is equalto 1. On the left, the order ord∞ f is an integer, and positive since f is a cusp form. Henceit follows from (7.3.1) that ord∞ f = 1 and that f is of order 0 at all points of H. In otherwords, f has no zeros in H. Of course we know, by example (7.1), that f is a multiple of�(z), and it is obvious from �(z) = η(z)24 that �(z) has no zeros.

Consider the Eisenstein series Gk for k ≥ 4. The functional equations, for γ = s andγ = u, are the following:

Gk(sz) = zkGk(z), Gk(uz) = zkGk(z).

The point i is a fixed point of s. Hence, by the first equation, if k �≡ 0 (mod 4), thenGk(i) = 0. Similarly, if k �≡ 0 (mod 6), then Gk(ρ) = 0. In particular,

G4(ρ) = 0, G6(i) = 0. (7.3.2)

It follows from (7.3.1) that the points in the orbit represented by ρ are the only zeros of G4and, moreover, that these points are simple zeros. Similarly, the functionG6 has simple zerosat the points in the orbit represented by i, and no other zeros.

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(7.4) Example. Klein’s j -invariant is the function defined in H by

j (z) := G4(z)3/�(z).

It is of weight 0. Since �(z) has no zeros in H, the function j (z) is holomorphic in H. It hasa simple pole at ∞ (since �(z) has a simple zero), with the Fourier expansion,

j (z) = q−1 + 744 + 196884q + · · · .The coefficients are integers. Indeed, it follows from (7.1.2) that �(z)−1 has an expansion�(z)−1 = q−1 +· · · with integral coefficients. As the Fourier coefficients ofG4 are integral,therefore, so are the coefficients of j (z).

The two values j (ρ) = 0 and j (i) = 1728 are immediate from (7.1.1) and (7.3.2).Consider, for λ ∈ C, the function f (z) = j (z)− λ and the equation (7.3.1). The right handside is 0, and the order at infinity is −1. It follows that exactly one of the remaining ordersis non-zero. Therefore, the function j (z) has the value λ exactly at the points of one singleorbit. In other words, Klein’s j -invariant defines a bijection,

H/ (1) ∼−→ C. (7.4.1)

In fact, it follows that j (z) has ρ as a triple zero, it takes the value 1728 with multiplicity 2 atthe point i, and it takes any other value with multiplicity 1 at the points of the correspondingorbit. It is easy to deduce that (7.4.1) is an analytic isomorphism.

(7.5) Note. The result of the previous example has as corollary the following theorem ofPicard: Any holomorphic function f (z) in C which avoids at least two values is constant.

Indeed, we may assume that f (z) avoids the two valuesw1 := j (ρ) = 0 andw2 := j (i) =1728. Let X by the open subset of H obtained by subtracting the two orbits containing i andρ, and let Y be the open subset of C obtained by subtracting to two points w1 and w2. Then,by (7.4), the j -invariant is an analytic isomorphism of X/ (1) onto Y . Since (1) actsproperly discontinuous on X, and without fixed points, it follows that j defines a coveringj : X → Y . By assumption, f maps C into Y . Therefore, since C is simply connected, f

lifts to a holomorphic map f : C → X. Consider the function eif (z). It is defined on C, andbounded, because f (z) takes values in the upper half plane; hence it is constant. It followsthat f (z) is constant. Therefore, f (z) is constant.

(7.6) Example. The field of (1)-automorphic functions is generated by Klein’s invariant,

M( (1)) = C(j). (7.6.1)

Indeed, the function j (z) has a simple pole at ∞ and no other poles. If u ∈ H, then thefunction j (z)− j (u) has a zero at u, and the (1)-order of the zero is equal to 1. It followsthat the function (j (z)− j (u))−m has a pole of orderm (i.e., a zero of (1)-order −m) at thepoints of the orbit represented byu, and no other poles. Letf be a non-zero (1)-automorphicfunction. It follows that by subtracting from f a linear combination of functions of the form,

jn, (j − j (u))−m,we can obtain a difference without poles. Thus the difference is a constant function. Thereforef belongs to C(j).

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[Autm] 43

(7.7) Example. Consider again the modular group (1). Let k be an arbitrary positive realnumber. By Proposition (3.4), a factor of weight k on (1) is completely determined by thesign at ∞ which is of the form,

ω∞ = e2πi(k−a)/12, (7.7.1)

where a is an integer, 0 ≤ a < 12. To get a homogeneous factor, assume that a is even.Denote by jk,a the corresponding factor, and set Mk(a) := M( (1), jk,a) etc. By (3.3), theparameters at ∞, i, and ρ of the factor jk,a are, respectively,

κ∞ = {(k − a)/12

}, κi = {

a/4}, κρ = {

a/3}.

Hence the number δk,a = δ( (1), jk,a) is given by the equation,

δk(a) = 1 + k/12 − {(k − a)/12

} − {a/4

} − {a/3

}.

When k is an even integer, the factor J k is obtained by taking a ≡ k (mod 12), and theexpression above reduces to that of Example (7.1).

Take k = 12 and a = 0. The corresponding factor of weight 1

2 is then the η-factor, seeExample (3.6). Clearly, δ1/2(0) = 1. Hence G( (1), jη) = S( (1), jη) is a 1-dimensionalspace generated by the η-function.

(7.8) Example. Consider the θ -group θ . The genus is g = 0, and μ = 12 since the index

of θ in (1) is equal to 3. For the θ -group, there is one elliptic orbit, represented by i oforder 2 with canonical generator γi = s. In addition, there are two cusps: one is representedby ∞, where the canonical generator is t2, and one is represented by −1 where the canonicalgenerator is conjugate to t . The θ -function belongs to G( θ , jθ ). The factor jθ has weightk = 1

2 , and the parameters at the three orbits are respectively,

κi = 0, κ∞ = 0, κ−1 = 18 .

Consequently, δ( θ , jθ ) = 1 + 12

12

12 − 1

8 = 1. Therefore, the space G( θ , jθ ) is 1-dimensional, and the θ -function is a generator. The θ -function is not a cusp form since it hasthe value 1 at ∞. Hence there are no non-trivial jθ -automorphic cusp forms, confirming thatδ′ = 0 since the cusp ∞ is jθ -regular and the cusp −1 is not.

The θ -function has a zero of order 1/8 at the cusp −1. Therefore, from the equation∑u ord θu f = 1/8, it follows that θ(z) has no zeros in H.Since θ(z) has no zeros in H and the value 1 at ∞, there is, for any real number l, a unique

determination of θ(z)l with the property that the limit of θ(z)l for z → ∞ is equal to 1.Obviously, θ(z)l is a j lθ -automorphic form, where j lθ is the factor of weight l/2 defined byj lθ (γ, z) = θ(γ z)l/θ(z)l . The factor j lθ has the following signs:

ωi(jlθ ) = 1, ω∞(j lθ ) = 1, ω−1(j

lθ ) = e2πil/8.

(The first two equations are obvious, the third can be seen from (2.7.3); of course it is obviouswhen l is an integer.) For convenience, take l = 2k where k is positive real. Then j 2k

θ is a

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factor of weight k on θ , and the corresponding parameters are respectively 0, 0, and{k/4

}.

It follows thatdim G( θ , j 2k

θ ) = 1 + k4 − {

k4

} = 1 + [k/4

].

The cusp ∞ is always regular, the cusp −1 is regular if and only if k ∈ 4Z. Hence,

dim S( θ , j 2kθ ) =

{ [k/4

] − 1 if k ∈ 4Z,[k/4

]otherwise.

Note that, if k ∈ 4Z, then j 2kθ = J k .

(7.9) Exercise. Prove the results of Example (7.1), using only the Special Case (6.6) ofTheorem B. [Hint: Estimate the dimension dim Gk for k = 2, . . . , 12. Use Equation (7.3.1)to show that the dimension is 0 for k = 2 and use the Eisenstein series to conclude that thedimension is 1 for k = 4, 6, 8, 10. Use the Eisenstein series and�(z), say defined by (7.1.1),to conclude that the dimension is 2 for k = 12. Investigate zeros and poles of �(z), andconclude that there is, for k ≥ 0, an exact sequence,

0 −→ Gk�−→ Gk+12 −→ C −→ 0.

Now deduce the results of Example (7.1).]

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[Autm] 45

8. The Proofs.

(8.1). Keep the setup of (6.1). In the calculations below we will need the formula,

∑u∈U

ordu f = 1

2πi

∫∂U

f ′(z) dzf (z)

. (8.1.1)

In the formula, U is an open subset of C with compact closure, f is a non-zero meromorphicfunction defined in an open domain containing the closure of U . The integral is the pathintegral over the oriented boundary of U , assumed to be sufficiently regular. It is assumedthat the function f has no zeros or poles on the boundary ∂U . The formula is a well knownconsequence of Cauchy’s residue formula.

The path integral on the right hand side of (8.1.1) can be formed over any path C that liesin the domain of f and avoids the zeros and poles of f . We define

IC(f ) := 1

2πi

∫C

f ′(z) dzf (z)

. (8.1.2)

The notation extends linearly to a chain C, defined as a formal integral linear combination oforiented paths. Note the following two equations,

IC(f1f2) = IC(f1)+ IC(f2), IφC(f ) = IC(f ◦φ). (8.1.3)

The first equation follows because the logarithmic derivative of the product function is thesum of the two logarithmic derivatives. In the second equation, the map φ is a holomorphicmap from some domain into the domain of f . The equation follows from the definition ofthe path integral, noting that d(φ(α(t)) = φ ′(α(t))dα(t) when φ is holomorphic and α is aC∞-map of the real variable t .

(8.2) Proof of Theorem B. We have to prove, for a non-zero j -automorphic form f , thefollowing formula: ∑

u mod

ord u f = kμ

2. (8.2.1)

The main observation used in the proof is the following: Let C be a path in D avoiding thezeros and poles of f and let γ be a matrix of . Then the following equation holds:

IγC(f ) = IC(f )+ kIC(Jγ ). (8.2.2)

Indeed, f ◦γ = jγ f since f is j -invariant. Hence, by (8.1.3), the left side is equal toIC(f )+IC(jγ ). Moreover, j (γ, z) = εJ (γ, z)k for some constant ε and some determinationJ (γ, z)k . Hence the logarithmic derivative of j (γ, z) is equal to k times the logarithmicderivative of J (γ, z). Thus IC(jγ ) = kIC(Jγ ).

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Let F be a finite normal fundamental domain for , and denote by U the interior of F .The domain F has a finite number of sides and vertices. Consider an infinite vertex u of F .By (Discr.3.14), u is a cusp for . Therefore, since f is exponentially bounded at the cusps, itfollows that f is holomorphic and non-zero in some fundamental neighborhood of u. Whena small fundamental neighborhood of each cusp of F is cut away from F , there remains iscompact set. As a consequence, the function f has only a finite number of zeros or poles inF (and in particular on the boundary of F ).

A side ofF can be divided into two by adding to the vertices a point of a side and the imageof the point under the boundary transformation corresponding to the side. Hence we mayassume that all zeros or poles of f on the boundary of F are finite vertices of F . In addition,we may assume that no side is mapped onto itself under the boundary transformation.

Let F0 be the subdomain obtained by cutting away from F a small fundamental neighbor-hood of each vertex. Choose the neighborhoods such that if V is the chosen neighborhoodof a vertex u and if γ u belongs to F , then γV is the chosen fundamental neighborhood ofγ u. There is only a finite number of zeros and poles of f in U . So, when the fundamentalneighborhoods are chosen sufficiently small, then all zeros and poles in U belong to the in-terior U0 of F0. The points of U are -ordinary, and hence the -order of f at a point u ofU0 is the ordinary order. Therefore, by (8.1.1),

∑u∈U

ord u f = I∂U0(f ). (8.2.3)

We will prove the formula (8.2.1) by evaluating carefully the integrals on the right side overthe various components of ∂U0.

The boundary ∂U0 has three types of components: there are line segments left from thesides ofF when the neighborhoods of the two end points are cut away, there are arcs consistingof the parts in F of the geodesic circles bounding the fundamental neighborhoods of the finitevertices, and there are segments consisting of the parts in F of the horo circles boundingthe fundamental neighborhoods of the infinite vertices of F . The contributions to the pathintegral in (8.2.3) coming from the components are grouped as follows: Denote by Isides(f )

the sum of the contributions coming from the sides. Choose in each -equivalence class ofvertices of F one vertex u, and denote by Iu(f ) the sum of the corresponding contributionsfrom the class; it is the sum of the path integrals along the parts in F of the boundaries of allfundamental neighborhoods of the vertices -equivalent to u. Accordingly,

I∂U0(f ) = Isides(f )+∑u

Iu(f ), (8.2.4)

where the sum is over the chosen system of representatives of the -equivalence classes ofvertices of F . Now, by (8.2.3) and (8.2.4), to prove Theorem B, it suffices to prove that, asthe neighborhoods shrink around the vertices,

Isides(f ) → kμ

2, Iu(f ) → − ord u f . (8.2.5)

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[Autm] 47

Consider first the contribution coming from the sides of F . For each side L of F thereis a component C = CL of ∂U0 lying on L and a corresponding component C ′ lying on theside γLL, where γL is the boundary transformation corresponding to L. The orientation ofC ′ is the reverse of the orientation of γLC. Hence, by (8.2.2), the two sides L and L′ = γLL

contribute to Isides(f ) with the sum,

IC(f )+ IC′(f ) = IC(f )− IγLC(f ) = −kIC(JγL).

Therefore,Isides(f ) = −k

∑′ICL(JγL),

where the sum is over unordered pairs {L,L′} of sides of F with L′ = γLL. Clearly, thesum on the right hand side converges, when the fundamental neighborhoods shrink aroundthe vertices, to the following sum

∑′IL(JγL) = 1

2πi

∑′∫L

J (γL, z)′

J (γL, z)dz .

The latter sum is, by (Discr.5.9.4), equal to −μ/2. Therefore, Isides(f ) converges to kμ/2,and the first relation in (8.2.5) has been proved.

Consider next the contribution Iu(f ) coming from a the vertices -equivalent tou. Choose(finitely many) matrices γi in so that the vertices -equivalent to u are the points ui = γiu.The fundamental neighborhoods of the points ui are of the images γiV of a fundamentalneighborhood V of u. Hence the components around the ui are of the form γiDi where theDi are segments of the boundary of V . The contribution Iu(f ) is then the sum,

∑i IγiDi (f ).

Let D be the union of the Di . Then, by (8.2.2),

Iu(f ) = ID(f )+ k∑

IDi (Jγi ). (8.2.6)

The terms in the sum on the right side converge to zero when the neighborhoods V shrinksaround u. Indeed, the assertion is obvious if u is a finite vertex, because then the integrandis bounded and the length of the integration path goes to 0. If u is an infinite vertex, we mayassume that the disk is the upper half plane and that u = ∞; in this case the assertion is easilyverified.

Therefore, to prove the second relation in (8.2.5), it suffices to prove the following equation,

ID(f ) = − ord u (f ). (8.2.7)

To prove (8.2.7), assume first that u is an infinite vertex. We may, after conjugation,assume that (D, u) = (H,∞). Then the canonical generator γu is a translation z �→ z + h.The fundamental neighborhood V is a half plane: Im z > R, and the components Di arehorizontal straight line segments on the boundary: z = R; they are oriented from the rightto the left. As observed in (Discr.3.16), the union D is a system of representatives for the

127


action of the canonical generator γu on the line z = R. Set q(z) := e2πiz/h. It follows fromthe local analysis in (5.1) that there is an equation,

f (z) = e2πiκuz/hg(q), (8.2.9)

where κu = κu(j) is the parameter and g is meromorphic in the unit disk: |q| < 1, say oforder N at the origin. By the choice of fundamental neighborhood V , the function f has nozeros or poles in the closed half plane z ≥ R. The half plane is mapped by q onto a pointeddisk: 0 < |q| ≤ ε. Hence g has no zeros or poles in the closed disk: |q| ≤ ε except possiblyat the origin. The image qD is the full boundary: |q| = ε of the disc, clockwise oriented.Therefore, by (8.1.1) and (8.1.3), ID(g◦q) = −N . The logarithmic derivative of the factore2πiκz/h is equal to 2πiκ/h. As D is the union of horizontal line segments, oriented fromright to left, of lengths adding up to h, it follows the ID(e2πiκuz/h) = −κu. Therefore, by(8.2.9),

ID(f ) = −κu −N = − ord u f,

and thus (8.2.7) holds.Assume next that u is a finite vertex. We may, after conjugation, assume that (D, u) =

(E, 0). Then the canonical generator γu is a rotation z �→ e2πi/euz. The fundamentalneighborhood V is a disk: |z| < ε, and the components Di are arcs on the boundary:|z| = ε; they are clockwise oriented. As observed in (Discr.3.16), the unionD is a system ofrepresentatives for the action of the canonical generator γu on the circle. Set w(z) := zeu . Itfollows from the local analysis in (5.3) that there is an equation

f (z) = zκueug(w), (8.2.10)

where κu = κu(j) is the parameter and g is meromorphic in the unit disk: |w| < 1, sayof order N at the origin. By the choice of fundamental neighborhood V , the function fhas no zeros or poles in the pointed disk: 0 < |z| ≤ ε. The pointed disk is mapped byw onto a pointed disk with radius εeu . Hence g has no zeros or poles in the image disk:|w| ≤ εeu except possibly at the origin. The image wD is the full boundary of the imagecircle |w| = εeu , clockwise oriented. Therefore, by (8.1.1) and (8.1.3), ID(g◦w) = −N .The logarithmic derivative of the factor zκueu is equal to κueuz−1. As D is the union ofarcs, clockwise oriented, of angles adding up to 2π/eu, it follows that ID(zκueu) = −κu.Therefore, by (8.2.10),

ID(f ) = −κu −N = − ord u f,

and thus (8.2.7) holds.Hence (8.2.7) holds in both cases, and the proof of Theorem B is complete.

(8.3). The proofs of Theorems A, C, D, E assume familiarity with the theory of Riemannsurfaces. Let X be a compact (connected) Riemann surface. It is well known that the onlyglobal holomorphic functions on X are the constants. Denote by M = M(X) the field ofmeromorphic functions on X. If ϕ �= 0 is a meromorphic function on X, then ϕ has finite

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[Autm] 49

order ordu ϕ at every point u of X. The order is zero except for a finite number of points u,and the sum of the orders is equal to 0:∑

u∈Xordu ϕ = 0. (8.3.1)

A divisor D on X is a finite formal sum, D = ∑nu.u, of points u of X. The coeffi-

cients ordu(D) := nu are integers, and equal to 0 except for a finite number of points u.The degree of the divisor is the sum, degD = ∑

ordu(D), of the coefficients. To everynon-zero meromorphic function ϕ there is associated a principal divisor, divϕ, defined byordu(divϕ) = ordu(ϕ), that is, by

divϕ =∑

(ordu ϕ).u.

It follows from (8.3.1) that the degree of a principal divisor is equal to 0,

deg(div ϕ) = 0. (8.3.2)

Two divisorsD and D ′ are called linearly equivalent, if the difference D −D′ is a principaldivisor. Thus linearly equivalent divisors have the same degree.

Associate with a given divisor D the following vector space over C of rational functionson X:

H 0(D) := {ϕ | divϕ +D ≥ 0},where the inequality divϕ+D ≥ 0 for divisors means the corresponding inequalities ordu ϕ+ordu D ≥ 0 for all the coefficients. Note that the inequality ordu ϕ+n ≥ 0 for n < 0 requiresϕ to have a zero at u of order at least −n, and for n ≥ 0 allows ϕ to have a pole of order at mostn at u. In particular, a function ϕ inH 0(D) is holomorphic except possibly at the finitely manypoints u where ordu(D) > 0. IfD andD′ are linearly equivalent, sayD−D′ = divψ , then,clearly, the multiplication, ϕ �→ ψϕ, defines a C-linear isomorphismH 0(D) ∼−→H 0(D′).

By definition, if ϕ is a non-zero function in H 0(D), then ordu ϕ + ordu(D) ≥ 0. Bytaking the sum over u, it follows that deg(divϕ)+ degD ≥ 0. Hence, by (8.3.2), degD ≥ 0.Therefore,

H 0(D) = 0 if degD < 0. (8.3.3)

Let ω be a non-zero meromorphic differential form. Locally, around a point u of X, wehave that ω = f dz, where z is a local parameter at u. By definition, the order ordu ω is theorder of f at u. Associate with ω the divisor,

divω =∑

(ordu ω).u.

Any meromorphic differential form ω′ is of the form ω′ = ϕω with a meromorphic functionϕ. It follows that divω′ = divϕ + divω. Hence the divisors of differential forms form onesingle class of divisors modulo principal divisors, called the canonical class. Any divisor inthe canonical class, that is, any divisor of a differential form,

K = divω,

is called a canonical divisor. Obviously, div(ϕω) = (div ϕ) + (divω). It follows that themap ϕ �→ ϕω induces an isomorphism fromH 0(K) to the space of holomorphic differentialforms.

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The Riemann–Roch Theorem. The vector spaces H0(D) are of finite dimension, and theyvanish if degD < 0. If degD ≥ 0, then the dimension ofH 0(D) is a most equal to degD+1.Moreover, there exists a number g = g(X) such that for any canonical divisor K and anydivisor D,

dimH 0(D) = degD + 1 − g + dimH 0(K −D). (8.3.4)

The number g = g(X) is called the genus of the Riemann surface X. It can be shown tobe equal to the topological genus of X as a surface.

Corollary. The genus g = g(X) is equal to the dimension of the space of holomorphicdifferential forms, and the degree of a canonical divisor is equal to 2g − 2,

g = dimH 0(K), degK = 2g − 2. (8.3.5)

Moreover, if degD ≥ 2g − 1, then dimH 0(D) = degD + 1 − g.

Proof. The first equality follows from (8.3.4) by taking D := 0. Next, the equality degK =2g−2 follows by takingD := K in (8.3.4). Finally, if degD ≥ 2g−1, then deg(K−D) < 0.Hence the last assertion of the Corollary follows from (8.3.3).

It follows from the Theorem and the Corollary that if degD ≥ 0, then

degD + 1 − g ≤ dimH 0(D) ≤ degD + 1,

and the first inequality is an equality if degD ≥ 2g − 1. The latter result is referred to asRiemann’s part of the Theorem.

(8.4). To apply the theory of Riemann surfaces to automorphic forms, note that the quotientX = D/ is, by the construction given in the proof of Corollary (Discr.2.13), a connectedRiemann surface. To get a local parameter around a point of X, choose a representative u inD ∪ ∂ D. Assume first that representative u is in D. After a suitable conjugation, we mayassume that (D, u) = (E, 0). Then the map,

w = zeu,

where the canonical generator at 0 is z �→ e2πieuz, identifies, for a small fundamental neigh-borhood U of u, the open subset U/ u of X with a small neighborhood of 0 in C; thus w isa local parameter around the given point ofX. Assume next the u is in ∂ D. After a suitableconjugation, we may assume that (D, u) = (H,∞). Then the map,

q = e2πiz/h,

where the canonical generator at ∞ is z �→ z + h, identifies, for a small fundamentalneighborhood U of u, the open subset U/ u of X with a small neighborhood of 0 in C; thusq is a local parameter around the given point of X.

Since is assumed to be a Fuchsian group of the first kind, the Riemann surfaceX is evencompact.

By construction, the field M( ) of -automorphic functions is the field of meromorphicfunctions onX. Moreover, the -order at u of a non-zero automorphic function ϕ is equal tothe order of ϕ as a meromorphic function on X.

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[Autm] 51

(8.5) Proof of Theorem A. As is well known, a compact Riemann surface is algebraic and, asa consequence, its field of meromorphic functions is finitely generated and of transcendencedegree 1 over C. Therefore, since X = X( ) is compact, Theorem A holds.

(8.6). Let X = X( ) be the Riemann surface of (8.4). Denote by Ccusp the cuspidal divisorof X, defined as the sum,

Ccusp :=∑

u a cusp1.u .

Associate with a non-zero j -automorphic form f the following divisor on X:

Df :=∑u

[ord u f

].u,

where [x] is the integral part. Recall that the -order is congruent to the parameter κu(j);whence

[ord u f

] = ordu f − κu(j). Therefore, by Theorem B, we obtain the equation,

degDf = kμ

2−

∑u

κu(j). (8.6.1)

As a consequence,degDf + 1 − g = δ(j). (8.6.2)

Fix the non-zero j -automorphic form f . By (4.9)(1), the multiplication ϕ �→ ϕf is anisomorphism M → M( , j). Under the multiplication, the product ϕf is an integral form,if and only if, for all u, ord u (ϕf ) ≥ 0, that is, if and only if the following inequality holds:

ordu ϕ + ord u f ≥ 0. (8.6.3)

The order ordu ϕ is an integer. Hence the inequality (8.6.3) is equivalent to the followinginequality:

ordu ϕ + [ord u f

] ≥ 0. (8.6.4)

It follows that ϕf is in G( , j) if and only if ϕ is in H 0(Df ). In other words, multiplicationby the fixed form f induces an isomorphism,

H 0(Df )∼−→G( , j). (8.6.5)

In particular, dimH 0(Df ) = dim G( , j).Clearly, if ϕf is an integral form, then ϕf is a cusp form, if and only if the inequality

(8.6.3) is strict for all cusps u. If a cusp u is j -irregular, then the parameter κu(j) is non-zero,and hence the order ord u f is not an integer; hence, the inequality in (8.6.3) is strict, if andonly if the inequality (8.6.4) holds. If a cusp u is j -regular, then the inequality in (8.6.3) isstrict if and only if the inequality (8.6.4) is strict, that is, if and only if,

ordu ϕ + [ord u f

] − 1 ≥ 0. (8.6.6)

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Hence ϕf is a cusp form if and only if (8.6.6) holds at all j -regular cusps u and (8.6.4) holdsat all other orbits. Let Cj -reg denote the j -regular part of the cuspidal divisor Ccusp, that is,Cj -reg = ∑

1.u where the sum is over the j -regular cusps u. It follows that ϕf is cusp formif and only if ϕ belongs toH 0(Df −Cj -reg). In other words, multiplication by the fixed formf defines an isomorphism,

H 0(Df − Cj -reg)∼−→S( , j). (8.6.7)

In particular, dimH 0(Df − Cj -reg) = dimS( , j).Obviously, the degree of Cj -reg is equal to the number of j -regular cusps. Hence, by

(8.6.2),deg(Df − Cj -reg)+ 1 − g = δ′(j). (8.6.8)

(8.7) Proofs of Theorems C and D. The assertions follow from Riemann’s part of the Riemann–Roch theorem, given the expressions (8.6.2) and (8.6.8) for δ(j) and δ ′(j), and the isomor-phisms (8.6.5) and (8.6.7).

(8.8). To use the full strength of the Riemann-Roch Theorem, we have to identify a canonicaldivisorK onX. Let f be a non-zero -automorphic form of weight 2, that is, f ∈ M( , J 2).Clearly, the differential form f dz on D is -invariant. Hence it descends to a meromorphicdifferential form on D/ , and in fact, as the following calculation shows, to a meromorphicdifferential form on X.

Consider first a point of D/ , represented by a point u of D. After conjugation, we mayassume that (D, u) = (E, 0). Then w = zeu is a local parameter at u. There is a normalizedLaurent expansion around u,

f (z) = zκueug(w),

and the parameter κu = κu(J2) is equal to 1 − 1/eu. Thus f (z) = (z/w)g(w); as dw =

euzeu−1dz = eu(w/z)dz, it follows that euf (z)dz = g(w)dw. Hence the order at u of

f (z)dz as a differential form on X is equal to the order of g(w) at 0. On the other hand, the -order ord u f is equal to κu plus the order of g(w). Therefore,

ord u f = ordu(f dz)+ (1 − 1eu). (8.8.1)

Consider next a cusp represented by a point u of ∂ D. After conjugation, we may assumethat (D, u) = (H,∞). Then q = e2πiz/h is a local parameter at u. There is a normalizedFourier expansion around u, and it has the form,

f (z) = g(q),

since the parameter κu = κu(J2) is equal to 0. As dq = (2πi/h)qdz, it follows that

(2πi/h)f (z)dz = q−1g(q)dq. Hence the order at u of f (z)dz as a differential form onX isequal to the order of g(q) minus 1. On the other hand, the -order of f is equal to the orderof g(q). Therefore, at a cusp u,

ord u f = ordu(f dz)+ 1. (8.8.2)

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[Autm] 53

If 1/eu is interpreted as 0 at a cusp u, then (8.8.2) is simply (8.8.1). The divisor Kf :=div(f dz) is a canonical divisor on X. From the two equations (8.8.1) and (8.8.2), it followsthat

Df = Kf + Ccusp. (8.8.3)

Take the degree in (8.8.3) and apply (8.6.1) with j := J 2 to obtain the equation,

μ = 2g − 2 +∑(

1 − 1

eu

). (8.8.4)

The equation is identical to (6.1.1). However, (6.1.1) was obtained with the topological genusand (8.8.4) was obtained with the holomorphic genus of X. Hence the two genera are equal,that is, the topological genus is equal to the dimension of the vector space of holomorphicdifferential forms.

In addition, since all cusps are J 2-regular, the isomorphism H 0(K) ∼−→S2( ) followsfrom (8.8.3) and (8.6.7). Hence the space S2( ) of cusp forms of weight 2 is isomorphic tothe space of holomorphic differential forms on X; in particular, dim S2( ) = g.

(8.9) Theorem H. Let j be the factor defined by j = J 2/j . Then a cusp u is j -regular if andonly if it is j -regular, and δ(j)+ δ(j ) is equal to the number of j -regular cusps. Moreover,the following equation holds,

dim G( , j) = δ(j)+ dimS( , j ). (8.9.1)

Proof. The factor j is of weight 2 − k and j j = J 2. Fix two non-zero automorphic forms,f ∈ M( , j) and f ∈ M( , j). Then the product f f belongs to M( , J 2). Hence thedivisorK := div(f f dz) is a canonical divisor. Now, for any point u,

ord u (f f ) = ord u f + ord u f . (8.9.2)

The fractional parts of the three orders in (8.9.2) are the parameters, κu(J 2), κu(j), andκu(j ). Hence either κu(J 2) = κu(j) + κu(j ) or κu(J 2) + 1 = κu(j) + κu(j ). Assumethat u is in D. Then κu(J 2) = 1 − 1/eu. Moreover, any parameter at u is, by (3.3.1), atmost equal to 1 − 1/eu. Therefore, κu(J 2) = κu(j)+ κu(j ). Assume next that u is a cusp.Then κu(J 2) = 0. If u is j -regular, then κu(j) = 0; it follows that also κu(j ) = 0 and thatκu(J

2) = κu(j)+ κu(j ). If u is j -irregular, then it follows that u is also j -irregular and thatκu(j)+ κu(j ) = 1. Hence κu(J 2)+ 1 = κu(j)+ κu(j ).

Now take integral parts of the orders in (8.9.2), and consider the corresponding divisors.On the left we obtain, by (8.8.3), the divisor K + Ccusp. On the right we obtain, by thediscussion above, the divisorDf +D

f+Cj -irreg. As a consequence, we obtain the equation,

K + Cj -reg = Df +Df. (8.9.3)

It was seen in the discussion above that a cusp u is j -regular if and only if it is j -regular. Takedegrees in (8.9.3) and use (8.3.5) and (8.6.2) to see that δ(j)+ δ(j ) is equal to the number ofj -regular cusps. Finally, apply (8.6.5) to (f, j) and (8.6.7) to (f , j ). By the Riemann-RochTheorem and (8.9.2), the final assertion of the Mail Theorem is a consequence.

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(8.10) Note. The factor j introduced in the Theorem may be called the dual factor. Sinceδ(j)+ δ(j ) is equal to the number of j -regular cusps, it follows that

δ(j ) = −δ′(j). (8.10.1)

The Main Theorem implies the Theorems C, D, and E. To obtain Theorem E, note first thatJ 2 is the dual to the trivial factor j = 1. The constant function f = 1 is -automorphicof weight 0, and the corresponding divisor is D1 = 0. Hence, δ(1) = 1 − g and sincedimH 0(0) = 1, the equation dim S2( ) = g follows from (8.9.1).

Note next that the factor J is self-dual. Hence δ(J ) is equal to half the number of regularcusps of , cf. (6.14.3). Therefore, the last part of Theorem E is a consequence of (8.9.1).Note that, more generally, any factor of weight 1 is of the formχ(γ )J (γ, z), and the dual is thefactor χ(γ )J (γ, z). In particular, if the character χ is quadratic, then the factor χ(γ )J (γ, z)is self dual.

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[Poinc] 1

Poincaré Series and Eisenstein Series

1. Poincaré series.

(1.1). Fix a finite disk D, a discrete subgroup of SL(D), and a homogeneous factor j on of real weight k. Assume that is a Fuchsian group of the first kind.

(1.2) Definition. Recall that the factor j defines an right action of on meromorphic functionsφ on D, determined by (φ ·j γ )(z) = j (γ, z)−1φ(z). Clearly, the series,∑

γ∈ φ ·j γ,

is formally -invariant. In general, of course, the series can not be expected to be convergent.Assume more generally that� is a given subgroup of and that φ is (�, j)-invariant, that

is, φ(γ z) = j (γ, z)φ(z) for all γ ∈ �. Consider the series,

G( , j,�, φ) =∑

γ∈�\ φ ·j γ, (1.2.1)

where the sum is over a system of representatives for the right cosets of � in . Since φ is(�, j)-invariant, the term in the sum corresponding to γ depends only on the coset containingγ . Again, the series is formally -invariant. The series (1.2.1) is the general Poincaré series.

The main questions associated with Poincaré series are the following:

(1) Is the series normally convergent in D? If it is, then the sumG(z) is meromorphic inD, and ( , j)-invariant. Moreover, if the given invariant function φ is holomorphicin D, then G(z) is holomorphic in D.

(2) Is G(z) exponentially bounded at the cusps of ? If it is, then G(z) is a ( , j)-automorphic form. If the given function φ is holomorphic, is G(z) then in integralform, or even a cusp form?

(3) Finally, is G(z) not the zero function?

(1.3). Consider an isomorphism α : D′ → D from a finite disk D′ onto D, defined by amatrix α in SL2(C). Recall that α defines a conjugate factor jα on the conjugate group α .Moreover, for every meromorphic function φ on D, there is a weight-k conjugate functionφα on D′ defined by φα(z′) = εJ (α, z′)kφ(αz′) with a fixed complex sign ε and a fixeddetermination of J (α, z)k. Clearly, if φ is (�, j)-invariant, then the conjugate function φα is(�α, jα)-invariant, and

G( , j,�, φ)α = G( α, jα,�α, φα). (1.3.1)

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[Poinc] 2 Automorphic functions16. februar 1995

(1.4) Definition. Let u be a point of D ∪ ∂ D and take the isotropy group u as �. Natural( u, j)-invariant functions are obtained as follows.

Assume first that (D, u) = (E, 0). The canonical generator γu at u = 0 is a diagonalmatrix, and the associated Möbius transformation is a rotation z �→ e2πi/euz. The functionj (γu, z) is constant, and equal to the sign e2πiκu , where κu = κu(j) is the parameter. Henceγu acts on functions on E by (φ ·j γu)(z) = e−2πiκuφ(e2πi/euz). It follows that the ( u, j)-invariant functions are the functions of the form φ(z) = zκueu φ(w), where w = zeu . Inparticular, take φ(w) = wl where l is an integer. Then φ(z) = z(κu+l)eu . Therefore, if κ isany real number congruent to the parameter κu(j) modulo Z, then the following function is( u, j)-invariant:

φ ,0,κ (z) = φκ(z) = zκeu . (1.4.1)

Assume next that (D, u) = (H,∞), where ∞ is -parabolic. The canonical generator γuat u = ∞ is an upper triangular matrix with ±1 in the diagonal, and the associated Möbiustransformation is a translation z �→ z + h. The function j (γu, z) is constant, and equalto the sign e2πiκu , where κu = κu(j) is the parameter. Hence γu acts on functions on H by(φ ·j γu)(z) = e−2πiκuφ(z+h). It follows that the ( u, j)-invariant functions are the functionsof the form φ(z) = e2πiκuz/hφ(q), where q = e2πiz/h. In particular, take φ(q) = ql wherel is an integer. Then φ = e2πi(κu+l)z/h. Therefore, if κ is any real number congruent to theparameter κu(j) modulo Z, then the following function is ( u, j)-invariant:

φ ,∞,κ (z) = φκ(z) = e2πiκz/h. (1.4.2)

In general, if u ∈ D∪∂ D, and κ is any real number congruent to the parameter κu(j)moduloZ, define a u-invariant function φu,κ as follows: in the two cases u ∈ D and u ∈ ∂ Drespectively, choose a Möbius transformation α : (D, u) → (E, 0) and α : (D, u) → (H,∞).Set

φu,κ := (φκ)α,

where the right hand side is the weight-k conjugate of the function defined for the conjugategroup α in (1.4.1) and (1.4.2) in the two cases respectively.

The Poincaré series G( , j, u, φu,κ) is denoted G( , j, u, κ), that is,

G( , j, u, κ)(z) =∑

γ∈ u\

φu,κ (γ z)

j (γ, z). (1.4.3)

In terms of the chosen conjugation α,

G( , j, u, κ)(z) = ε∑ φκ(αγ z)

J (α, γ z)kj (γ, z). (1.4.4)

It should be emphasized that the Poincaré series (1.4.3) is only defined when κ is congruentto the parameter κu(j) of j at u. If u is a -ordinary point, then the canonical generator γuis the identity, and the parameter κu(j) is equal to 0. The parameter is also zero, when u is aj -regular cusp. Hence, in these two cases, the series is defined for integer values of κ . Theseries (1.4.3) for a j -regular cusp u and κ = 0 is called the Eisenstein series associated withthe cusp u.

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[Poinc] 3

(1.5) Note. In spite of the notation on the left hand side of (1.4.3), the series does dependon choices. First, there is a sign involved in the definition of the weight-k conjugate under α(of course, when k is an integer, then there is a unique k’th power J (α, z)k, and we can takeε = 1). But the function φu,κ = (φκ)

α , and hence also the Poincaré series, depends on thechoice of α.

Compare the function φu = φu,κ obtained from α with the function φ ′u obtained from a

second choice α′. The isomorphisms α and α ′ differ by an automorphism δ of the target,α′ = δα. By definition, φu = (φ)α , where φ is the function (1.4.1) or (1.4.2) respectivelyobtained from the conjugate group α . Similarly, φ′

u = (φ′)δα where φ′ is obtained from theconjugate group α′

. Hence it suffices to compare φ and (φ′)δ .Assume first that u is in D. Then δ is a rotation in E around 0, say δz = ζ z with |ζ | = 1.

The two functions φ and φ ′ are the same function zκeu . Moreover, up to a complex sign,the conjugate function φ ′δ is equal to φ(ζz) = (ζ )κeuφ(z) and hence, up to sign, equal to φ.Therefore, the function φ′

u is, up to sign, equal to φu. It follows that the series (1.4.3) for apoint u ∈ D is well defined up to a complex sign.

Assume next that u is a -parabolic point. Then δ is a Möbius transformation in H of theform δz = rz+ b with r > 0 and a real number b. The two functions φ and φ ′ are different:φ = e2πiκz/h where h is defined from α and φ′ = e2πiκz/h′

where h′ is defined from α′ .

Clearly, h′ = rh. Hence, up to the sign e2πiκb/(rh), the function (φ′)(δz) is equal to φ(z).The weight-k conjugate φ′δ is obtained from φ ′(δz) by dividing by J (δ, z)k. The modulus ofJ (δ, z)k is equal to r−k/2. Hence φ′δ is equal to φ multiplied with a nonzero complex number(of modulus rk/2). Therefore, the function φ′

u is, up to multiplication by a nonzero number,equal to φu. It follows that the series (1.4.3) for a -parabolic point u is well defined only upto multiplication by a non-zero number.

From (1.3.1) we obtain, for a general conjugation, the following equation up to multipli-cation by a nonzero number:

G( , j, u, κ)α = G( α, jα, α−1u, κ). (1.5.1)

(1.6) Example. Take (D, , j) = (H, (1), J k) where k is an even integer. The cusp ∞ isJ k-regular since k is even. Hence the Eisenstein series Gk = G( (1), J k,∞, 0) is defined.The group is homogeneous and the isotropy group at ∞ is dicyclic, generated by the matrixt . Obviously, the map, [

a b

c d

]�→ (c, d),

induces a bijection from ∞\ onto the set of pairs modulo ±1 of relatively prime integers(c, d). Thus the Eisenstein series corresponding to the cusp ∞ and κ = 0 is the following,

Gk(∞, 0)(z) = 1

2

∑(c,d)=1

1

(cz + d)k.

In other words, the series is the normalized Eisenstein seriesGk(z). It is normally convergentin H for k ≥ 4.

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(1.7) Theorem G. Assume that the weight k of the factor j is greater than 2. Let u be apoint of D∪ ∂ D, and let κ be a number congruent modulo Z to the parameter κu(j). Then,in the setup of (1.4), the Poincaré series G = G( , j, u, κ) defines a ( , j)-automorphicform. The functionG(z) is holomorphic at all points of D that are not -equivalent to u. Inaddition, it vanishes at all -parabolic points that are not -equivalent to u. At the givenpoint u, the -order of G is nonnegative if κ ≥ 0; if κ < 0, then the -order is equal to κ .Finally, if κ = 0 and u is a -parabolic point (necessarily j -regular), then G(z) does notvanish at u.

Proof. By (1.5.1), we may, after conjugation, assume that the disk is the upper half planeH. In general, we think of the Poincaré series (1.4.1) as indexed by a chosen system ofrepresentatives γ for the cosets u\ . The term corresponding to γ is the function,

gγ (z) = φu,κ (γ z)/j (γ, z),

and the Poincaré series is the series G(z) = ∑γ gγ (z).

The proof to the Theorem will be given in three parts below. In part I, we prove that theseries is normally convergent in H. It follows that the series defines a meromorphic ( , j)-invariant function G(z) in H. By construction, the function φu is holomorphic in H, exceptpossibly when u ∈ H and κ < 0. In the exceptional case, the term gγ (z) in the series has apole of order −κ at γ−1u.

Therefore, to finish the proof of the Theorem, it remains to study the behavior of G(z)at the -parabolic points: we have to prove that G(z) is exponentially bounded and that theremaining assertions of the theorem about -parabolic points hold. Clearly, to study thebehavior of G(z) at a given -parabolic point we may, after conjugation, assume that thegiven -parabolic point is the point ∞. Thus, in parts II and III, we assume that the point ∞is -parabolic.

The possible poles for the terms gγ (z) are the points in the orbit u (when u ∈ H andκ < 0). Therefore, since ∞ is -parabolic, there is a number R > 0 so that all the termsare holomorphic in the half plane HR : z > R. Hence G(z) is holomorphic in HR . In partII, the series G(z) is broken up into partial sums Gw(z) as follows: The group acts on theright on the index set u\ . In particular, the subgroup ∞ acts. Hence, the index set is splitinto disjoint ∞-orbits. Accordingly, the series G(z) is split into a sum of partial series,

G(z) =∑

Gw(z), (1.7.1)

where the sum is over all ∞-orbits w of u\ , and Gw(z) is the series,

Gw(z) =∑

gγ (z),

where the sum is over those representatives γ for which the right coset uγ belongs to theorbit w. As a partial sum of the series G(z), each series Gw(z) is normally convergent inH; in particular, the function Gw(z) is holomorphic in HR . Moreover, the series (1.7.1) isnormally convergent. As a sum over a ∞-orbit, each function Gw(z) is ( ∞, j)-invariant.

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[Poinc] 5

Consider the ∞-orbit w containing a given right coset uβ where β is in . Clearly,under the right action of ∞, the isotropy group of uβ is the intersection,

∞ ∩ β−1 uβ = ∞ ∩ β−1u.

A matrix in the intersection has ∞ and β−1u as fixed points. Hence, the intersection is thetrivial subgroup 1 or ±1 unless ∞ = β−1u. Of course, the exceptional case is only possibleif u is -parabolic and -equivalent to ∞. In the exceptional case, where β∞ = u, the orbitw consists of the single right coset uβ; the corresponding partial series will be denotedGu.It consists of a single term,

Gu(z) = gβ(z) where β∞ = u. (1.7.2)

In all other cases, the partial series corresponding to the orbit w containing uβ is the fol-lowing:

Gw(z) =∑

γ∈P ∞gβγ (z). (1.7.3)

In part II we prove that each partial series Gw(z), which is not the exceptional series(1.7.2), converges to 0 for z → ∞. The latter result is then used to finish the proof of theTheorem in part III.

(1.8) Part I of the proof. It will be convenient to consider the measure in H defined bydHk (z) = yk/2−2dxdy (where z = x + iy). The measure dH0 (z) = y−2dxdy is SL2(R)-invariant. It follows that the measure dHk (z) respects weight-k conjugation as follows: if α isa matrix of SL2(R) and g is continuous on the subset αM of H, then

∫M

|gα(z)|dHk (z) =∫αM

|g(z)|dHk (z). (1.8.1)

To prove that the Poincaré series converges normally, let K be a compact subset of H.Choose δ > 0 and a compact subset M of H such that, for any w in K , the closed disk:|z−w| ≤ δ is contained inM . LetC be the constant defined byC := (πδ2)−1 max(z)2−k/2where the maximum is over z ∈ M . For any function g(z) holomorphic in a neighborhoodof M we have, for w ∈ K , the equation,

πδ2g(w) =∫

|z−w|≤δg(z)dxdy. (1.8.2)

Indeed, in the disk, g(z) is the uniform limit of its power series expansion around w. Hence,using polar coordinates to evaluate the integral

∫(z − w)ndxdy, it suffices to note that∫ 2π

0 einθdθ = 0 for n > 0. Thus the equation (1.8.2) holds. From the equation, we ob-tain the estimate πδ2|g(w)| ≤ ∫

M|g(z)|dxdy. Hence, by the definition of C,

|g(w)| ≤ C

∫M

|g(z)|dHk (z). (1.8.3)

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The term gγ (z) in the Poincaré series is equal to φ(γ z)/j (γ, z) where φ = φu,κ . Hencegγ (z) is holomorphic inH except possibly for a pole in the point γ −1u if u ∈ H. The compactset M meets only a finite number of points in the orbit u. Therefore, except for a finitenumber of terms, the term gγ (z) is holomorphic in a neighborhood of M . The term gγ isequal to φ ·j γ , and in particular, it is a weight-k conjugate of φ(z). Therefore, by (1.8.1) and(1.8.3), we have, except for a finite number of terms in the series, the estimate for w ∈ K ,

|gγ (w)| ≤ C

∫γM

|φ(z)|dHk (z). (1.8.4)

Hence, to prove that the Poincaré series converges normally, it suffices to prove the followingassertion: the sum of the integrals in (1.8.4), over the representatives γ for which u /∈ γM ,is finite.

To the latter assertion will be proved for an arbitrary compact subset M of H. We mayassume that M is contained in a fundamental domain for . Indeed, if F is a fundamentaldomain for , thenM meets only a finite number of transforms of F . Hence M decomposesinto a finite number of pieces each of which is contained in a fundamental domain and, clearly,if the assertion holds for each piece, then it holds for M .

SinceM is contained in a fundamental domain for , the sum of the integrals in (1.8.4) isequal to the integral over the union,

M ′ :=⋃

γM,

where the union is over the representatives γ of the cosets u\ for which u /∈ γM . Thus itsuffices to prove that the following integral is finite:

∫M ′

|φ(z)|dHk (z). (1.8.5)

Assume first that u ∈ H. A small neighborhood of u meets only a finite number oftransforms of a fundamental domain, and hence only a finite number of transforms γM . Itfollows that the unionM ′ is contained in the subdomain H′ obtained from H by cutting awaya small neighborhood of u. Hence, it suffices to prove that the following integral is finite:

∫H′

|φ(z)|dHk (z).

The function φ is, by definition, the weight-k conjugate obtained from φ0,κ = zeuκ by aMöbius transformation α : (H, u) → (E, 0). Clearly, a suitable Möbius transformation is themap: α(z) = (z − u)/(z − u). Hence, up to a constant,

φ(z) = (z− u)−kφ0,κ (αz).

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[Poinc] 7

The function φ0,κ is bounded in the complement of a neighborhood of 0, and so the functionφ(z) is bounded in H′. Hence, it suffices to prove that the following integral is finite:

∫H′

1

|z − u|k yk/2−2dxdy.

The finiteness follows easily using polar coordinates: the integral∫ ∞εr−k+k/2−2rdr is finite

because k > 0, and the integral∫ π

0 (sin θ)k/2−2dθ is finite, because k > 2.

Assume next that u is -parabolic. Clearly, after a conjugation, we may assume thatu = ∞. Let z �→ z + h be the Möbius transformation associated to the canonical generatorγ∞. The union M ′ is over a system of representatives for the cosets ∞\ . Since M iscompact and ∞ is -parabolic, there is a number R so that no transform γM intersects thehalf plane: z > R. Consider the part of M ′ in the vertical strip: nh ≤ z ≤ (n+ 1)h. Thepart is transformed, by the power γ −n∞ , into the vertical strip 0 ≤ z ≤ h. Moreover, sinceφ(z) is ( u, j)-invariant, it follows from (1.8.1) that the integral over the part is unchangedwhen the part is replaced by its transform. Therefore, we may assume thatM ′ is contained inthe part H′ of H determined by the inequalities 0 < z ≤ R, 0 ≤ z ≤ h. Hence, it sufficesto prove that the following integral is finite,

∫H′

|φ(z)|dHk (z).

The function φ = φ∞,κ is, by definition, the function

φ(z) = e2πiκz/h.

It is bounded in the horizontal strip 0 < z ≤ R. In particular, it is bounded in H ′. Hence, itsuffices to prove that the following integral is finite:

∫H′yk/2−2dxdy.

The finiteness is obvious: the integral∫ R

0 yk/2−2dy is finite, because k > 2.Thus it has been proved in both cases that the integral (1.8.5) is finite. Hence the Poincaré

series converges normally in H, and part I of the proof is complete.

(1.9) Part II of the proof. By part I of the proof, we know that the Poincaré series G(z)is a ( , j)-invariant meromorphic function in H. Hence Theorem G holds if there are no -parabolic points. In the remaining parts of the proof we assume that ∞ is -parabolic. Letz �→ z+ h be the Möbius transformation associated to the canonical generator at ∞. In thissecond part of the proof, we prove that the partial series Gw(z) of (1.7.3), where β ∈ andβ∞ �= u, converges to 0 for z → ∞.

The term gβγ in the series (1.7.3) is equal to (φu,κ ·j β) ·j γ . The function φu,κ is aweight-k conjugate of either φ0,κ or φ∞,κ , and φu,κ ·j β is a weight-k conjugate of φu,κ .

141


Therefore, replacing u by β−1u, we may assume that β = 1. Then u �= ∞. Therefore, forthe conjugation α defining φu,κ = (φκ)

α , we have that J (α, z) = cz + d where c �= 0. Infact, if u ∈ H we may assume that J (α, z) = z − u and if u ∈ ∂ H we have c and d in R.Hence, dividing by a nonzero number, we may assume that J (α, z) = z + d, where d hasnonnegative imaginary part. It follows that |φu,κ | = |z+ d|−k |φκ(αz). The function φκ(αz)is equal to (αz)κeu if u ∈ H and equal to e2πiκ(αz)/h if u ∈ ∂ H. Since u �= ∞, it followsin both cases that the function φκ(αz) is bounded in HR. Hence, with a constant C > 0, wehave the estimate for z ∈ HR:

|φu,κ (z)| ≤ C|z + d|−k.The translation z �→ z+h associated to γ∞ generates the group P ∞. Moreover |J (γ n∞, z)| =1. Therefore, for z ∈ HR ,

|Gw(z)| ≤ C∑n

1

|z + nh+ d|k .

Clearly, the sum on the right side converges to 0 uniformly as z → ∞.Thus we have proved that the partial seriesGw(z), except in the exceptional case, converges

to 0 uniformly as z → ∞, and part II of the proof is complete.

(1.10) Part III of the proof. Assume as in part II that the point ∞ is -parabolic and thatP ∞ is generated by z �→ z + h. We have to study the behavior of G(z) near ∞. It sufficesto consider three cases:

(1) The point u is in H.(2) The point u is -parabolic and not -equivalent to ∞.(3) The point u is equal to ∞.

We have to prove in all cases that G(z) is exponentially bounded at ∞. In fact, we have toprove in the cases (1) and (2) that the order ofG at ∞ is positive; in case (3) we have to provethat the -order is positive if κ > 0, and equal to κ is κ ≤ 0.

By part II of the proof, the functionG(z) is a normally convergent seriesG(z) = ∑Gw(z)

of functions Gw(z) holomorphic in a half plane HR . One of the functions Gw(z) may bethe exceptional function Gu of (1.7.2). Denote by

∑′w Gw(z) the series of the remaining

functions. Each function Gw(z) is ( ∞, j)-invariant, and hence of the form,

Gw(z) = e2πiκ∞z/hGw(q), q = e2πiz/h,

where G(q) as a function of q is meromorphic in the punctured unit disk: 0 < |q| < 1. Thehalf planeHR is mapped by q onto a punctured diskV−0 whereV is an the open disk: |q| < ε.Since Gw(z) is holomorphic in HR , the function Gw(q) is holomorphic in the pointed diskV −0. Moreover, sinceGw(z) → 0 for z → ∞ by part II and κ∞ < 1, it follows that Gw(q)is also holomorphic for q = 0; in addition, Gw(q) vanishes for q = 0 when κ∞ = 0. As theseries

∑′Gw(z) converges normally in HR , it follows that the series

∑′Gw(q) converges

normally in the pointed disk V − 0. Moreover, each termGw(q) in the series is holomorphic

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[Poinc] 9

also at q = 0. Therefore, as is well known, the series∑′

Gw(q) converges normally in thewhole disk V . In particular, the sum G(q) := ∑

Gw(q) is holomorphic in V ; in addition, itvanishes at q = 0 if κ∞ = 0. Therefore, since

∑′Gw(z) = e2πiκ∞z/hG(q),

it follows that the series∑′

w Gw(z) has positive order at ∞.Now, in the cases (1) and (2), we have G(z) = ∑′

w Gw(z). Hence the assertions for thecases (1) and (2) hold. In case (3), the function G(z) is the sum of two functions,

G(z) = Gu(z)+∑′

Gw(z).

As u = ∞, the first function is φ∞(z) = e2πiκz/h, and so its -order at ∞ is equal to κ . Thesecond function has positive -order. Clearly, the assertion for the case (3) is a consequence.

Thus part III of the proof is complete, and Theorem G has been proved.

(1.11) Note. Consider, in the assumptions of Theorem G, the series G(z) = G( , j, u, κ)

for κ ≥ 0. It follows from the Theorem that the form G(z) is an integral form. If u ∈ D,then G(z) is a cusp form and if u ∈ ∂ D, then G(z) is a cusp form if an only if κ > 0. Notehowever that the formG(z) may be the zero form. In fact, the examples in (Autm.7) containseveral cases where S( , j) = 0 (for k > 2). In these cases, necessarily G(z) = 0.

It follows from the Theorem that the Eisenstein series G( , j, u, 0) associated with a j -regular -parabolic point u is an integral non-vanishing form in G( , j). It has order 0 atthe cusp defined by u, and it vanishes at all other cusps. It follows that the evaluation map in(Autm.6.13) is surjective. In particular, we recover the result that the codimension of S( , j)in G( , j), for k > 2, is equal to the number of j -regular cusps.

(1.12) Note. It is a consequence of Theorem G that there are non-zero ( , j)-automorphicforms for a factor j of any weight k. Indeed, assume first that k > 2. Then the Poincaréseries G(z) = G( , j, u, κ) obtained from, say, a -ordinary point u and κ < 0 has -orderequal to κ . In particular, G(z) is a nonzero function in M( , j). For general k, choose aneven integer l > 2 such that k + l > 2. Then there are nonzero functions f ∈ M( , jJ l)

and g ∈ M( , J l). Whence the quotient f/g is a nonzero function in M( , j).

143


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[Poinc] 11

2. A Fourier expansion of an Eisenstein Series.

(2.1). In this section, the disk is assumed to be the upper half plane H and k is assumed to bean integer. We consider a level-N subgroup of the modular group (1) = SL2(Z) and on a factor on of the following form,

j (γ, z) = J (γ, z)k/χ(γ ), (2.1.1)

where χ : → C∗ is a unitary character. We make a series of assumptions:(i) By hypothesis, is the preimage in (1), under the reduction map modulo N , of a

subgroup of SL2(Z/N). We assume that χ is a level-N character on , that is, χ is thecomposition of the reduction map and a character χ : → C∗.

(ii) We assume that contains the subgroup of diagonal matrices in SL2(Z/N). Equiv-alently, since the subgroup of diagonal matrices corresponds to the subgroup 0

0(N), weassume that ⊇ 0

0(N).(iii) It follows from the assumption in (ii) that is homogeneous. In order that the factor

(2.1.1) is homogeneous, we assume that

χ(−1) = (−1)k. (2.1.2)

(iv) The -parabolic points consist of the rational numbers and the point ∞, since isof finite index in (1). Since is homogeneous, the canonical generator at ∞ is the matrixth for some positive integer h. We assume that the point ∞ is a j -regular cusp, that is, weassume that

χ(th) = 1. (2.1.3)

(2.2). Denote by + the unipotent subgroup of , that is, + is the intersection of andthe subgroup of upper triangular matrices with 1 in the diagonal. Then + is the cyclicsubgroup generated by γ∞, and it is of index 2 in ∞. Clearly, the image of + in is theunipotent subgroup + of . By assumption (2.1)(iv), the character χ : → C∗ is trivial onthe subgroup +.

For any 2 by 2 matrix β, denote by 2β the second row of β. Clearly, two matrices γ1 andγ2 with determinant 1 have the same second row if and only if γ1γ

−12 is unipotent. It follows

in particular that the map γ �→ 2γ defines an injection,

+\ ↪→ Z2.

The image of the map will be denoted 2 . It consists of the pairs (c, d) of integers that occuras the second row of some matrix of . Similarly, there is an injection,

+\ ↪→ (Z/N)2.

The image 2 consists of pairs of residue classes modulo N that occur as second row of amatrix of . The character χ : → C∗ is trivial on the subgroup +, and hence it induces a

145


map from the set +\ to C∗. Given the bijection above, the latter map may be viewed as amap from 2 to C∗; we extend it with the value 0 to a map from (Z/N)2 to C. The extendedmap,

χ : (Z/N)2 → C,

defines, by composition with the reduction map modulo N , a map χ : Z2 → C. For α ∈SL2(Z), denote by

αχ : Z2 → C

the map obtained by composition of χ : Z2 → C and right multiplication by α. Unwindingthe definition, the value αχ(c, d) for integers c and d is determined as follows: if there existsa matrix γ of such that (c, d) modulo N is the second row of γα, then αχ(c, d) = χ(γ );otherwise αχ(c, d) is equal to 0.

By assumption (2.1)(ii), the group contains the subgroup of diagonal matrices ofSL2(Z/N). It follows that the row (0, n) belongs to 2 if and only if n is invertible in Z/N .Moreover, the group (Z/N)∗ of invertible elements n is isomorphic to the group of diagonalmatrices of SL2(Z/N). Hence χ defines, by restriction, a character χ : (Z/N)∗ → C∗, andthe function χ(0, n) is obtained from the character by extending it with the value 0 on residueclasses that are not invertible.

The map χ(n) := χ(0, n) is a residue character modulo N . From (2.1.2) we obtain theequation,

χ(−n) = (−1)kχ(n). (2.2.1)

In addition, for integers c and d we have the equation,

αχ(nc, nd) = χ(n) αχ(c, d). (2.2.2)

Indeed, if n is not prime to N , then clearly both sides are equal to 0. Assume that n isprime to N , and let δn be a matrix of whose reduction moduloN is a diagonal matrix withlast row (0, n). If γ is a matrix of such that 2γ α = (c, d), then δnγ belongs to and2δnγ α = (nc, nd); hence

αχ(nc, nd) = χ(δnγ ) = χ(δn)χ(γ ) = χ(n) αχ(c, d).

Thus the equation (2.2.2) holds if the right hand side is non-zero. It follows, replacing n byits inverse modulo N , that it holds if the left hand side is non-zero. Hence (2.2.2) holds.

(2.3). Consider, for k ≥ 3, the Eisenstein series,

G(z) = G( , j,∞, 0)(z),

where j is the factor of (2.1.1). By Theorem G, the series is an integral j -automorphic form,and its order at ∞ is equal to 0. It vanishes at all cusps different from ∞. We will consider

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[Poinc] 13

the Fourier expansion of G(z) at any -parabolic point v. So, let α be a matrix of SL2(Z)such that v = α∞. Consider the weight-k conjugate series, Gα(z), that is,

Gα(z) =∑

γ∈ ∞\

1

J (α, z)kj (γ, αz)=

∑γ∈ ∞\

χ(γ )

J (γ α, z)k,

where the last equation follows from the definition of j and the automorphy equations forJ (γ, z). Each right coset modulo ∞ splits into two cosets modulo +. Hence, if we formthe sum over the cosets modulo +, each term is repeated twice, and we obtain the equation,

2Gα(z) =∑

γ∈ +\

χ(γ )

J (γ α, z)k.

By definition of the function αχ , the equation may be rewritten as follows,

2Gα(z) =∑

(c,d)∈2 α

αχ(c, d)

(cz + d)k. (2.3.1)

The set of pairs (c, d) in 2 α is equal to the set of pairs (c, d) such that c and d are relativelyprime and αχ(c, d) is non-zero. Indeed, obviously, if a pair (c, d) belongs to the first set, thenit belongs to the second set. Conversely, assume that (c, d) belongs to the second set. Sincec and d are prime, there is a matrix β in SL2(Z) such that 2β = (c, d). Since αχ(c, d) �= 0,there is a matrix γ in such that, moduloN , 2γα ≡ (c, d). The two matrices β and γα havemodulo N the same second row. Therefore, modulo N , the quotient (γ α)β−1 is a unipotentmatrix of SL2(Z/N). The latter matrix can be lifted to a unipotent matrix τ of SL2(Z).Replace β by τβ. The replacement does not change the second row, so (c, d) is the secondrow of the new β. Moreover, for the new β, the quotient (γ α)β−1 is modulo N equal to 1.Therefore, the quotient belongs to (N). Since (N) is contained in , it follows that thequotient (γ α)β−1 is in . Thus β = γ ′α with a matrix γ ′ of . As (c, d) = 2β, it followsthat (c, d) belongs to the first set 2 α.

It follows that in the sum (2.3.1) is unchanged if it is formed over all pairs (c, d) of relativelyprime integers. Consider the sum over all non-zero pairs (c, d) of integers. By grouping theterms according to the greatest common divisor, we obtain by (2.2.2) the equation,

∑′ αχ(c, d)

(cz + d)k=

∞∑n=1

χ(n)

nk

∑(c,d)=1

αχ(c, d)

(cz + d)k.

The third sum is, as noted above, equal to the sum in (2.3.1). Therefore, the following equationholds, ∑′ αχ(c, d)

(cz + d)k=

∞∑n=1

χ(n)

nk

(2Gα(z)

). (2.3.2)

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By (2.2.1), the right hand side of (2.3.2) is unchanged if the factor 2 is omitted and the sumover n ≥ 1 is replaced by the sum over n �= 0. Whence,

∑n�=0

χ(n)

nkGα(z) =

∑′ αχ(c, d)

(cz + d)k. (2.3.3)

Note that the factor∑

n�=0 χ(n)n−k is non-zero. Indeed, in the sum over n ≥ 1, the first term

is equal to 1, and the sum of the remaining terms is, in absolute value, at most∑

n≥2 n−k ≤∫ ∞

1 t−kdt = 1/(k − 1).It follows from (2.3.3) that the seriesGα(z), apart from the factor

∑n�=0 χ(n)n

−k , is equal

to the Eisenstein series Eαχk (z) considered in (App.2).

(2.4) Proposition. The Eisenstein seriesG(z) of (2.3) has at the cusp ∞ the Fourier expan-sion, with q = e2πiz/N ,

G(z) = 1 + 2

Ak(χ)

(2i

N

)k ∑r≥1

σχk−1(r)q

r . (2.4.1)

The constant Ak(χ) is the special number associated with the character χ(n), and σ χk−1(r)

is the weighted sum of (k − 1)’st powers of divisors,

σχk−1(r) =

∑d|r

∑a mod N

χ(r/d, a)e2πiad/Ndk−1. (2.4.2)

At a cusp v = α∞ which is not -equivalent to ∞, the Fourier expansion is the following,

Gα(z) = 2

Ak(χ)

(2i

N

)k ∑r≥1

σαχk−1(r)q

r , (2.4.3)

where the definition of σαχk−1(r) is analogous to (2.4.2).

Proof. By (App.1.10.2), for the constant in (2.3.3), we have the equation,

∑n�=0

χ(n)

nk= (−1)kπk

(k − 1)!Ak(χ). (2.4.4)

It follows from (2.1.2) that αχ(−c,−d) = (−1)k αχ(c, d). Hence, in the notation of(App.2), we have that

(αχ

)∗ = (−1)k αχ .Now, the series G(z) is given by (2.3.3) with α = 1. By definition, χ(n) = χ(0, n).

Hence the equation (2.4.1) follows directly from Corollary (App.2.5).Assume that v = α∞ is not -equivalent to ∞. If a matrix γα has (0, 1) as its second row,

then it has ∞ as fixed point; thus γ v = γα(∞) = ∞. Hence no matrix γα with γ ∈ has(0, 1) as its second row. Therefore, αχ(0, 1) = 0. It follows from (2.2.2) that αχ(0, n) = 0for all n. Consequently, the special number associated with the function αχ is equal to 0.Hence the equation (2.4.3) follows from the expansion for the Eisenstein series E

αχk (z) in

Corollary (App.2.5) by dividing by the constant (2.4.4).

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[Poinc] 15

(2.5) Note. The Fourier expansion of the Eisenstein series G(z) in (2.4) is in terms of theparameter qN = e2πiz/N . On the other hand, G(z) has period h, where th is the canonicalgenerator of at ∞, and so G(z) has a Fourier expansion in terms of qh = e2πiz/h. Clearly,the matrix tN belongs (N). Hence it belongs to . Consequently, tN is a power of t lh , that is,N = lh. It follows that the Fourier coefficient in (2.4.1) to qr vanishes unless r is a multipleof l.

(2.6) Note. The Eisenstein series is not normally convergent for k = 2. However, the series onthe right hand side of (2.3.3), when the summation is performed as in [App.2.2], is normallyconvergent, and if Gα(z) is defined by the equation (2.3.3) for k = 2, then the expansion of(2.4) holds. However, the resulting functionGα(z) can not be expected to be ( , j)-invariantin general.

When k = 1, the situation is even more complicated. In order that the summation describedin [App.2.2] applies to the right hand side of (2.3.3), it is required that the sum

∑dαχ(c, d)

over d modulo N is equal to 0.

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150


[Poinc] 17

3. Example: Eisenstein series for the theta group.

(3.1). Recall that the θ -group θ is the subgroup of (1) formed by matrices that modulo 2are congruent to either 1 or s. It is generated by the matrices t 2 and s. The θ -factor jθ is afactor of weight 1

2 . It is determined on the generators by the equations,

jθ (t2, z) = 1, jθ (s, z) =

√z

i.

Let k be an integer. Then the power j 2kθ is a factor of integral weight k. In particular, the

square j 2θ is a factor of weight 1, and hence of the form,

j 2θ (γ, z) = J (γ, z)/χθ(γ ),

where χθ : θ → C∗ is a unitary character. The character is given on the generators asfollows:

χθ (t2) = 1 χθ(s) = i.

It follows that the character χθ is the character χθ considered in Exercise (Mdlar.3.8). It isgiven as follows,

χθ

[a b

c d

]=

⎧⎪⎪⎪⎨⎪⎪⎪⎩

1 for d ≡ 1 (mod 4),

i for c ≡ 1 (mod 4),

−1 for d ≡ −1 (mod 4),

−i for c ≡ −1 (mod 4).

The group θ is a level-4 group, and it follows from the description the χθ is a level-4character. Clearly, the setup of (2.1) applies. The character χθ (n) is a Dirichlet charactermodulo 4, usually denoted χ4(n). The value χ4(n) is 1 if n ≡ 1, it is −1 if n ≡ −1, and it is0 if n is even. The function χθ (c, d) is determined by the expression above. On a nonzeropair (c, d), the value is given as follows:

χθ (c, d) =⎧⎨⎩χ4(d) if c is even and d is odd,

iχ4(c) if c is odd and d is even,

0 otherwise.

(3.2). Assume that k is at least 3. Then the results of Sections 1 and 2 apply to the Eisensteinseries,

Gk(z) := G( θ , Jk/χkθ ,∞, 0).

It follows that Gk(z) is an integral form in Gk( θ , χkθ ). Moreover, its Fourier expansion isgiven by (2.4). Let us fix k and write χ for χ k

θ . To determine the Fourier coefficients of theEisenstein series, we need the special number Ak(χ) and the function σχk−1(r). The result isthe following.

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(3.3) Proposition. The Eisenstein seriesGk(z) of (3.2), for k ≥ 3, has the following Fourierexpansion, with q = e2πiz/2,

Gk(z) = 1 +∑r≥1

βk(r)qr ,

where the coefficients βk(r) are given by the formula,

βk(r) = 4

Ak

[2k−1σ ′

k−1(r)+ σ ′′k−1(r)

];the Ak are the numbers of (App.1.3) and

σ ′k−1(r) =

∑d|r

χ4(r/d)kdk−1, σ ′′

k−1(r) =∑

d|r, d≡k mod 2

(−1)(d−k)/2dk−1.

Proof. The expansion is given by (2.4), however in terms of q4 = e2πiz/4. As Gk(z) isperiodic with period 2, we know a priori that the coefficients σ χk−1(r) are only non-zero whenr is even.

To determine the constant factor in the expansion (2.4.1), note that the function χ(n) =χ4(n)

k depends on the parity of k. Assume first that k is odd. Then χ(n) = χ4(n). Clearly,χ4 = 2δ−χ2 where δ(n) is the function introduced in Example (App.1.13). Hence,Ak(χ) =2Ak(δ) − Ak(χ2). It follows from (App.1.13) and (App.1.12), for k odd, that 2Ak(δ) =(−1)kAk/2k and Ak(χ2) = 0. Therefore,

Ak(χ) = (−1)kAk/2k. (3.3.1)

If k is even, then χ(n) is the function χ2(n) considered in Example (App.1.12); hence thespecial number is given by the formulaAk(χ) = Ak/2k . Therefore, the formula (3.3.1) holdsfor all k.

From (3.3.1) we obtain, for the constant factor in the expansion (2.4.1), the equation,(2/Ak(χ))(2i/4)k = 2(−i)k/Ak . Hence the expansion is the following,

Gk(z) = 1 + 4

Ak

∑r≥1

1

2ikσχk−1(r) q

r4 .

Therefore, we have to prove that σχk−1(r) vanishes when r is odd and that σ χk−1(2r) =2ikβk(r).

The sum σχk−1(r) is over divisors d of r and over a modulo 4. Denote by σ 0,2(r) the

sum of the terms corresponding to a = 0 and a = 2 and by σ 1,3(r) the sum of the termscorresponding to a = 1 and a = 3. Clearly, for the first sum,

σ 0,2(r) =∑d|r

[χ(r/d, 0)+ χ(r/d, 2)(−1)d

]dk−1 =

∑d|r

ikχ4(r/d)k[1 + (−1)d

]dk−1.

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[Poinc] 19

Obviously, the terms in the sum corresponding to odd divisors d vanish. In particular, whenr is odd, the sum is zero. For an argument of the form 2r , the nonzero terms are obtainedfrom the even divisors of 2r , that is, for divisors of the form 2d for d | r . Whence,

σ 0,2(2r) = 2ik∑d|r

χ4(r/d)k2k−1dk−1 = 2ik2k−1σ ′

k−1(r). (3.3.2)

For the second sum,

σ 1,3(r) =∑d|r

[χ(r/d, 1)id + χ(r/d, 3)(−i)d]dk−1 =

∑d|r, r

deven

[1kid + (−1)k(−i)d]dk−1.

Evidently, the sum is zero if r is odd. For an argument of the form 2r , the nonzero terms areobtained from the divisors d of r . Hence,

σ 1,3(2r) =∑d|r

id[1 + (−1)k+d)

]dk−1 = 2ik

∑d|r, d≡k

id−kdk−1 = 2ikσ ′′k−1(r). (3.3.3)

Since σχk−1(r) = σ 0,2(r)+ σ 1,3(r), it follows that σ χk−1(r) vanishes for odd r , and it followsfrom (3.3.2) and (3.3.3) that σχk−1(2r) = 2ikβk(r). Hence Gk(z) has the asserted Fourierexpansion.

(3.4) Note. Depending on the residue class of k modulo 4, the expression for βk(r) can besimplified as follows:

Assume first that k is even. Then χ4(n)k = χ2(n). Hence σ ′

k−1(r) = ∑′d|r dk−1 where the

prime indicates a sum over those divisors d for which r/d is odd. Clearly, σ ′′k−1(r) vanishes

for odd r , and for even r ,

σ ′′k−1(r) = (−1)k/22k−1

∑d| r2(−1)ddk−1.

The sum splits into a difference of two:∑ev

d| r2 dk−1 − ∑odd

d| r2 dk−1, where the two sums are,

respectively, over the even and odd divisors of r/2. It follows, for k even, that βk(r) is equalto 2k+1/Ak times the following expression,

∑′

d|rdk−1 + (−1)k/2

∑ev

d| r2dk−1 − (−1)k/2

∑odd

d| r2dk−1. (3.4.1)

If r is odd, then the last two sums vanish, and the first is equal to σk−1(r). Assume that ris even. Clearly, the last sum is the sum over the odd divisors d of r . The first sum is overthe (necessarily) even divisors d for which r/d is odd, and the second sum is over the evendivisors d for which r/d is even. Thus every divisor d of r contributes with a non-zero term

153


in exactly one of the three sums in the expression. It follows that the expression, for k ≡ 0(mod 4), is equal to σ ev

k−1(r)− σ oddk−1(r). Hence, for k ≡ 0 (mod 4),

βk(r) = 2k+1

Ak×

{σk−1(r) if r is odd,

σ evk−1(r)− σ odd

k−1(r) if r is even.

Similarly, for r even and k ≡ 2 (mod 4), the expression (3.4.1) is equal to the differenceσk−1(r)− 2

∑evd| r2 d

k−1. Now the even divisors of r2 are of the form 2d where d is a divisor

of r4 . Therefore, for k ≡ 2 (mod 4),

βk(r) = 2k+1

Ak×

{σk−1(r) if r is odd,

σk−1(r)− 2kσk−1(r/4) if r is even.

The difference in the last expression equals (1+2k−1)σ oddk−1(r)+ (4k−1 −2k)σk−1(r). Indeed,

if r is even, then a divisor in r is either odd, or it is of the form 2d where d is an odd divisorin r , or it is of the form 4d where d is a divisor of r

4 . Hence, σk−1(r) = (1 + 2k−1)σ oddk−1(r)+

4k−1σk−1(r4 ), and we obtain the alternative expression for the difference.

Assume that k is odd. Then χ4(n)k = χ4(n). The sum in σ ′′

k−1(r) is over the odd divisorsd of r . If d is odd, then id−k = (−1)(d−1)/2(−1)(k−1)/2. Moreover, χ4(d) = (−1)(d−1)/2.Therefore, for odd k,

βk(r) = 4

Ak

∑d|r

[2k−1χ4(r/d)+ (−1)(k−1)/2χ4(d)

]dk−1.

(3.5). The θ -function is (essentially) given by its Fourier expansion at the cusp ∞, that is,θ(z) = ∑

n qn2

where q = e2πiz/2. The θ -function belongs to G( θ , jθ ), and it vanishes atthe second cusp represented by −1. Hence the power θ(z)l , for a positive integer l, belongsto G( θ , j lθ ) and it vanishes at −1. Clearly, the power has the Fourier expansion at ∞,

θ(z)l = 1 +∑r≥1

bl(r)qr ,

where bl(r) is the number of solutions (n1, . . . , nl) in Zl to the equation,

r = n 21 + · · · + n 2

l .

Assume that l is even, say l = 2k, where k is a positive integer. Then j 2kθ = (j 2

θ )k = J k/χkθ .

Hence θ(z)2k belongs to the space G( θ , J k/χ kθ ) and it vanishes at −1. The Eisensteinseries Gk(z), for k ≥ 3, has the same properties. Moreover, both forms have at ∞ a Fourierexpansion with constant term 1. It follows that the difference θ(z)2k −Gk(z) is a cusp form.

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[Poinc] 21

(3.6) Proposition. Let b2k(r) be the number of solutions (n1, . . . , n2k) in Z2k to the equation,

r = n 21 + · · · + n 2

2k. (3.6.1)

Then, for k ≥ 3, the following asymptotic formulas holds:

b2k(r)− βk(r) = O(rk/2).

Moreover, the equality b2k(r) = βk(r), holds for k = 2, 3, 4, that is,

b4(r) = 8 ×{σ(r) when r is odd,

3σ odd(r) when r is even.

b6(r) = 4∑d|r

[4χ4(r/d)− χ4(d)

]d2.

b8(r) = 16 ×{σ3(r) when r is odd,

σ ev3 (r)− σ odd

3 (r) when r is even.

Finally, for k = 1, we have the equation, b2(r) = 12β1(r) = 4

∑d|r χ4(d).

Proof. The difference b2k(r)−βk(r), for k ≥ r , is the r’th Fourier coefficient in the cusp formθ(z)2k − Gk(z). The asymptotic formula, for k ≥ 3, follows from Theorem F (Autm.6.16)since the difference θ(z)2k −Gk(z) is a cusp form.

Moreover, for k ≤ 4, the space Sk( θ , j 2kθ ) of cusp forms is equal to 0 by Example

(Autm.7.8). Therefore, the equation b2k(r) = βk(r) holds for 3 ≤ k ≤ 4. It holds also fork = 2, because it is possible to prove that the Poincaré seriesG2(z)with a suitable summationdoes define an automorphic form. The proof is not easy, and it is not covered in these notes.

Finally, the explicit formula, b2(r) = 12

∑d|r χ4(r) is elementary. It can be proved using

unique prime factorization in the ring Z[i] of Gaussian integers.

(3.7) Exercise. StudyGk(z) at the cusp−1 using conjugation by the transformationu = st−1.Prove that

uχθ(c, d) =⎧⎨⎩iχ4(−d) if c and d are odd,

iχ4(c − d) if c is odd and d is even,

0 otherwise.

Deduce from (2.4.3) and (Autm.5.2) a formula for the number bodd2k (r) of decompositions of

r into 2k odd squares.

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Automorphic functions24. juni 2011

[App] 1

Appendix

1. Bernoulli and Euler numbers.

(1.1) Setup. As is well known, two important sequences of numbers, the Bernoulli numbersB0, B1, . . . and the Euler numbers E0, E1, . . . are defined by the Taylor expansions,

z

ez − 1=

∑k≥0

Bkzk

k!,

2ez

e2z + 1=

∑k≥0

Ekzk

k!. (1.1.1)

The first function has poles at the non-zero integral multiples of 2πi. Hence the first series isconvergent in the open disk: |z| < 2π . Similarly, the second series converges when |z| < π .Obviously, B0 = E0 = 1.

The numbers appear in the Taylor expansion of many important functions. The secondfunction in (1.1.2) is equal to 1/ cosh z. In particular, it is an even function of z. Hence allEuler numbers of odd index vanish. Moreover, evaluation at iz yields the expansion,

1

cos z=

∑k even

(−1)k/2Ekzk

k!. (1.1.2)

In the first equation of (1.1.1), divide by z and add 12 to obtain the following equation:

1

2

ez + 1

ez − 1= 1

z+ (B1 + 1

2 )+∑k≥1

1

k + 1Bk+1

zk

k!.

In the equation, the left side is an odd function of z. Therefore, the equation implies thatB1 = − 1

2 and that all other Bernoulli numbers of odd index vanish. The equation is theexpansion of the function 1

2coth z2 . Evaluate at 2iz and multiply by 2i to obtain,

cot z = 1

z+

∑k odd

(−1)(k+1)/22k+1

k + 1Bk+1

zk

k!. (1.1.3)

Finally, from cot z− 2 cot 2z = tan z, we obtain the expansion,

tan z =∑k odd

(−1)(k−1)/22k+1(2k+1 − 1)

k + 1Bk+1

zk

k!. (1.1.4)

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[App] 2 Automorphic functions24. juni 2011

(1.2) Definition. It will be convenient to introduce a third sequence of numbers A0, A1, . . . ,called the Euler zigzag numbers or up/down numbers. They are defined by the expansion,

α(z) = 1 + sin z

cos z=

∑k≥0

Akzk

k!. (1.2.1)

Obviously, A0 = 1. The function α(z) is the sum of the even function sec z = 1/ cos z andthe odd function tan z. Hence (1.2.1) implies the expansions,

1

cos z=

∑k even

Akzk

k!, tan z =

∑k odd

Akzk

k!. (1.2.2)

Accordingly, the Ak with even k are called secant numbers, and the Ak with odd k are calledtangent numbers.

By comparing (1.2.2) with the expansions (1.1.2) and (1.1.4) we obtain the followingequations for k ≥ 0:

Ak ={(−1)k/2Ek if k is even,

(−1)(k−1)/22k+1(2k+1 − 1)Bk+1/(k + 1) if k is odd,

or,

Ek = (−1)k/2Ak, k even, Bk = (−1)k/2−1k

2k(2k − 1)Ak−1, k ≥ 2 even.

(1.3) Lemma. The zigzag numbers Ak for k ≥ 0 are positive. whole numbers. Moreover,if k ≥ 2 is even, then kAk−1 is divisible by 2k−1. Furthermore, the numbers are given byA0 = 1 and the following recursion formula for k ≥ 0:

Ak+1 = Ak +(k

2

)Ak−1 −

(k

4

)Ak−3 +

(k

6

)Ak−5 ± · · · .

Proof. The function α(z) of (1.2.1) is the sum of sec z = 1/ cos z and tan z. To obtain thederivatives of α(z), note that (sec z)′ = sec z tan z and (tan z)′ = 1 + tan2 z. It followsby induction on k that there are polynomials F0(t), F1(t), . . . and G1(t),G2(t), . . . and anequation,

α(k)(z) = sec z Fk(tan z)+Gk+1(tan z).

The polynomials are given recursively: F0 = 1, G1 = t , and Fk+1 = tFk + (1 + t2)F ′k and

Gk+1 = (1+ t2)G′k . It follows easily that each ofFk andGk is a polynomial of degree k, with

positive integer coefficients in degrees k, k − 2, k − 4, . . . and zero coefficients otherwise.In particular, Ak = Fk(0)+Gk+1(0) is a positive integer. Hence the first part of the Lemmaholds.

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[App] 3

To prove the second part, consider the Taylor expansion,

τ(z) = 2z tan z =∑

Ckzk

k!.

By (1.2.2), if k ≥ 2 is even then Ck is the (even) integer 2kAk−1 and Ck vanishes otherwise.It suffices to prove that the integer Ck is divisible by 2k .

Clearly, τ ′(z) = 2 tan z + 2z(1 + tan2 z). Multiplication by 2z yields the equation,

2zτ ′(z)− 2τ(z) = 4z2 + τ(z)2.

It follows by comparing the coefficients of zk/k! for k > 2 that

2(k − 1)Ck =k∑i=0

(k

i

)CiCk−i , for k > 2. (1.3.2)

As noted above, Ck = 0 if k is odd, or if k = 0. Moreover, C2 = 4. Proceed by induction,and assume that k > 2 is even, k = 2l. In the sum (1.3.2), the extreme terms vanish, becauseC0 = 0. In the middle term, the binomial coefficient

(kl

)is even,

(kl

) = 2(k−1l−1

). The remaining

terms come in equal pairs, since(ki

) = (kk−i

). Therefore, division by 2 in (1.3.2) yields:

(k − 1)Ck =∑

0<i<l

(k

i

)CiCk−i +

(k − 1

l − 1

)C 2l . (1.3.3)

By the induction hypothesis, each product CiCk−i (including C 2l ) is divisible by 2k . So the

sum is divisible by 2k . Hence, so is Ck , because the factor k − 1 is odd.To prove the recursion formula, note that α(z) = α ′(z) cos z. Insert the expansionsα(z) =∑Akz

k/k! and α′(z) = ∑Ak+1z

k/k! and the well known expansion of cos z. The recursionformula follows by comparing the coefficients of zk.

(1.4). A simple computation using the recursion formula gives the following table:

k 0 1 2 3 4 5 6 7 8 9 10 11Ak 1 1 1 2 5 24 61 24 · 17 1385 28 · 31 50521 29 · 691

(1.5) Definition. Letψ : Z → C be a periodic function, sayψ(n+N) = ψ(n) for all integersn. Associate withψ the sequence of transformed Bernoulli numbersAk(ψ) for k ≥ 0 definedby the expansion,

αψ(z) =∑

b mod N

ψ(b)

Ncot

z− πb

N= 1

z

∑k≥0

Ak(ψ)zk

k!, (1.5.1)

where the index b ranges over a system of representatives for the residue classes modulo N .At z = 0, only the factor 1

Ncot(z/N) in the term for b = 0 contributes with a pole; it is a

simple pole with residue 1. Hence A0(ψ) = ψ(0). Clearly αψ(z) has period π , with polesat the numbers z = aπ . In particular, the series in (1.5.1) converges for |z| < π .

The relation between that numbers Ak(ψ) and the generalized Bernoulli numbers Bk(ψ)is described in (1.15).

159


(1.6). The starting point of the following computations is the well known formula,

1

z+

∑n≥1

( 1

z + n+ 1

z− n

)= π cot πz. (1.6.1)

The series on the left is normally convergent, and so it defines a meromorphic function in C.To prove the formula, note that the function on the left hand side has the following properties:it has poles exactly at the integers, with principal part around z = n equal to 1/(z − n), it isperiodic with period 1, and it is bounded in any domain |z| ≥ ε. Clearly, the function on theright has the same properties. It follows that the difference of the two functions is boundedin C, and hence constant. As the two functions are odd functions, the difference is zero.

We shall rewrite the formula as follows:∑+

n

1

z+ n= π cot πz,

where the summation sign∑+ indicates the symmetric summation over all integers n, that

is, the sum with the terms ordered as indicated in (1.6.1). Note that the sum∑

n(z+ n)−k isabsolutely convergent when k ≥ 2.

(1.7) Lemma. Let ψ : Z → C be a periodic map. Then the following formula holds:∑+

n

ψ(n)

z+ n= −παψ(−πz). (1.7.1)

Proof. Fix a system of representatives a for the residue classes modulo a period N . Thenthe symmetric summation on the left side of (1.7.1) can be replaced by the sum over a ofthe symmetric summation over n of the terms ψ(a)/(z + a +Nn). For fixed a we have, by(1.6.1), the equation,∑+

n

1

z+ a +Nn= 1

Nπ cot

πz+ πa

N= −π 1

Ncot

−πz− πa

N.

Multiply by ψ(a), and form the sum over a. Clearly, the result is the right side of (1.7,1).

(1.8) Proposition. Let ψ : Z → C be a periodic map, and form the function αψ(z) in (1.5.1).Then, for k ≥ 1, ∑+

n

ψ(n)

(z + n)k= −πk(k − 1)!

α(k−1)ψ (−πz). (1.8.1)

As a consequence, ∑+

n�=0

ψ(n)

nk= −πkAk(ψ)

k!. (1.8.2)

Proof. The first formula is obtained from (1.7.1) by applying the operator d k−1/dzk−1 andmultiplying the result by (−1)k−1/(k − 1)!. As is clear from the definition in (1.5.1), thefunction αψ(z) − ψ(0)/z has no pole at z = 0, and the value at z = 0 of its (k − 1)’stderivative is Ak/k for k ≥ 1. Hence, after subtracting from both sides of (1.8.1) the termψ(0)/zk, the formula (1.8.2) is obtained by evaluating the resulting equation at z = 0.

160


[App] 5

(1.9) Note. In the following examples it is convenient to obtain trigonometric relations usinglogarithmic derivation: the logarithmic derivative of F(z) is the derivative of logF(z), equalto F ′(z)/F (z). Logarithmic derivation turns multiplication into addition.

For instance, the function α(z) of (1.2.1) is equal to cos z/(1 − sin z), and hence equalto the logarithmic derivative of A(z) = 1/(1 − sin z). The function cot z is the logarithmicderivative of sin z. Consequently, the function αψ(z) is the logarithmic derivative of thefunction,

Fψ(z) :=∏

b mod N

(sin

z − πb

N

)ψ(b).

The exponents ψ(b) are not assumed to be integers, so Fψ(z) is a multi valued function.

(1.10) Example. For the function χ ≡ 1 (N = 1) we have α1(z) = cot z. So the correspond-ing transformed Bernoulli numbers Ak(1) are the Taylor coefficients of z cot z. They may beobtained from (1.1.3). Alternatively, with F1 = sin z and A(z) = (1 − sin z)−1 we have that

F1(2z)/F1(z) = 2 cos z = 2/√A(z)A(−z).

Take the logarithmic derivative and multiply by z:

(2z)α1(2z)− zα1(z) = −z2

(α(z)− α(−z)).

Therefore, for the k’th Taylor coefficient: (2k − 1)Ak(1) = −12 (1 + (−1)k)kAk−1, that is,

A0(1) = 1, Ak(1) = −k2k − 1

Ak−1, if k ≥ 2 even,

A0(1) = 1, and Ak(1) = 0 otherwise.Hence, by (1.8.2), if k ≥ 2 is even, then 2ζ(k) = ∑

n�=0 n−k = πkAk−1/(2k − 1)(k− 1)!.

Equivalently,

1

1k+ 1

2k+ 1

3k+ · · · = πkAk−1

2(2k − 1)(k − 1)!, k ≥ 2 even. (1.10.1)

(1.11) Example. Let ψ = ψ1/2 be the parity character defined by ψ1/2(n) = 1 when n isodd and ψ1/2(n) = 0 if n is even (N = 2). The transformed Bernoulli numbers Ak(ψ1/2)

are determined from F1/2 := sin( z2 − π2 ). Obviously,

F1/2(2z) = − cos z = −1/√A(z)A(−z).


(2z)α1/2(2z) = −z2

(α(z)+ α(−z)).

161

[App] 6 Automorphic functions16. marts 2006

Therefore, by taking the k’th Taylor coefficient,

Ak(ψ1/2) = −kAk−1

2kwhen k ≥ 2 is even,

and Ak(ψ1/2) = 0 otherwise. Hence, by (1.8.2),

1

1k+ 1

3k+ 1

5k+ · · · = πkAk−1

2k+1(k − 1)!, k ≥ 2 even. (1.11.1)

(1.12) Example. Letψ1/4 be the characteristic function of 1 modulo 4, defined byψ1/4(n) = 1if n ≡ 1 (mod 4) and ψ1/4(n) = 0 otherwise (N = 4). The transformed Bernoulli numbersAk(ψ1/4) are determined from F1/4 = sin( z4 − π

4 ). Clearly,

2F1/4(2z)2 = 2 sin2( z2 − π

4 ) = 1 − cos(z− π2 ) = 1 − sin z = A(z)−1.


2(2z)α1/2(2z) = −zα(z).

Hence Ak(ψ1/4) = −kAk−1/2k+1. As a consequence, for all k ≥ 1, we have the formula,

∑+

n≡1 (mod 4)

1

nk= πkAk−1

2k+1(k − 1)!. (1.12.1)

If k is odd, it follows that

1

1k− 1

3k+ 1

5k± · · · = πkAk−1

2k+1(k − 1)!, k ≥ 1 odd. (1.12.2)

and if k is even, we obtain the result from Example (1.11).

(1.13) Exercise. Prove for S(z) = 2 sin z the formula,

S(Nz) =N−1∏j=0

S(z + jπ/N).

[Hint: useXN −1 = ∏N−1j=0 (X− e2πij/N).] Deduce that the transformed Bernoulli numbers

Ak(ψ) are independent of the choice of period N entering in their definition.

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Automorphic functions16. marts 2006

[App] 7

(1.14) Exercise. Prove thatAk is equal to the number of up-down permutations of (1, . . . , k),that is, permutations (σ1, . . . , σk) with σ1 < σ2 > σ3 < σ4 > σ5 < · · · .

Hint: From the equation 2α ′(z) = α(z)2 + 1, deduce the recursion formula,

2Ak =k−1∑i=0

(k − 1

i

)AiAk−i−1, A0 = A1 = 1.

Prove for the number ak of up-down permutations the following formulas (for k ≥ 1):

ak =k−1∑i=0i odd

(k − 1

i

)aiak−i−1 =

k−1∑i=0i even

(k − 1

i

)aiak−i−1.

The i’th terms in the two sums count the numbers of up-down permutations having, respec-tively, the biggest element k at position i + 1 and the smallest element 1 at position i + 1.

(1.15) Exercise. The transformed Bernoulli numbers Ak(ψ) introduced here are closelyrelated to the generalized Bernoulli numbers Bk(ψ) of Leopoldt, defined by the followingexpansion,

βψ(z) =N∑a=1

ψ(a)zeaz

eNz − 1=

∞∑k=0

Bk(ψ)zk

k!.

Indeed, let ψ−(n) = ψ(−n) and let ψ be the discrete Fourier transform of ψ , defined by theequation,

ψ(n) =∑

ψ(a)ζ an, ζ = e2πi/N .

Then the following equation holds:

αψ(z) = − iNψ(0)+ 2i

Nβψ−(

2izN).

Prove the equation and deduce the following equation by comparing coefficients to zk−1:

Ak(ψ) = ( 2iN

)kBk(ψ−), k �= 1, A1(ψ) = −i

Nψ(0)+ 2i

NB1(ψ−).

Prove for a primitive multiplicative character χ : (Z/N)∗ → C∗, extended to all of Z withthe value zero on integers n not prime to N , that

χ = τ(χ, ζ )χ,

where τ is the Gauss sum.

(1.16) Exercise. Prove for k even that kAk/2k−1 is odd. [Hint: Induction. Use (1.3.3)divided by 2k . Rewrite the sum using the equation

(ki

) = (k−1i−1

) + (k−1i

). The resulting

expression should not depend on the parity of l. Deduce modulo 2 that the sum equals∑1≤j<l

(k−1j

) = 2k−2 − 1.]

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[App] 8 Automorphic functions16. marts 2006

164


[App] 9

2. Fourier expansion of Eisenstein series.

(2.1) Setup. Fix a natural number N , let ζ = ζN be the canonical N ’th root of unity,ζ := e2πi/N , and let q = qN be the function of z defined by q(z) = e2πiz/N . In addition, letτ(ψ, ξ) be the Gauss sum, defined for any N ’th root of unity ξ as the sum,

τ(ψ, ξ) :=∑b

ψ(b)ξb, (2.1.1)

over a system of representatives b of the residue classes modulo N .In Section 1 we studied the function

∑+ψ(n)/(z + n) where the sum is the symmetric

summation. We proved the equation∑+

ψ(n)/(z + n) = −παψ(−πz), and we used it tostudy the function around z = 0. By definition of the function αψ(z) in (1.5), the equation isthe following: ∑+

n

ψ(n)

z+ n= π

N

∑ψ(b) cot

π(z + b)

N. (2.1.2)

To study the function in the upper half plane H, use the equation,

cot z = ieiz + e−iz

eiz − e−iz= i

(1 − 2

1 − e2iz

).

With z := π(z + b)/N it follows that

cotπ(z + b)

N= i

(1 − 2

1 − ζ bq

), ζ = e2πi/N , q = e2πiz/N .

For z ∈ H we have |q| < 1, and hence 1/(1 − ζ bq) = ∑d≥0 ζ

bdqd . Therefore, from (2.1.1)and (2.1.2) we obtain the expansion,

∑+

n

ψ(n)

z+ n= πi

N

∑b

ψ(b)− 2πi

N

∑d≥0

τ(ψ, ζ d)qd , z ∈ H. (2.1.3)

Apply the operator dk−1/dzk−1 and multiply by (−1)k−1/(k − 1)!. Since q ′(z) =(2πi/N)q(z), we obtain, for k ≥ 2 and z ∈ H, the formula,

∑n

χ(n)

(z + n)k= (−1)k

(k − 1)!

(2πi

N

)k ∞∑d=1

τ(χ, ζ d)dk−1qd . (2.1.4)

(2.2) Definition. Let χ : Z × Z → C be a function of 2 integer variables, and set χm(n) :=χ(m, n). Assume that the function χ is bounded. In addition, assume for all m in Z thatχm(n), as a function of n, has period N . Consider for k ≥ 1 the associated Eisenstein series,

Eχk (z) :=

∑′ χ(m, n)

(mz+ n)k, (2.2.1)

165


where the sum is over (m, n) �= (0, 0). The series is normally convergent for k ≥ 3 anddefines a holomorphic function in the upper half plane H. If the summation is arranged asfollows:

Eχk (z) :=

∑n�=0

χ(0, n)

nk+

∑m�=0

∑n

χ(m, n)

(mz+ n)k, (2.2.2)

then the series is normally convergent also for k = 2 and defines a holomorphic functionEχ2 (z). In fact, as we shall show below, if the function χ satisfies that

∑a mod N χ(m, a) = 0

for allm, and if the summations overn in (2.2.2) are interpreted as the symmetric summations,then the right hand side of (2.2.2) is even convergent for k = 1.

For every natural number r , consider the following linear combination of k’th powers ofthe divisors of r:

σχk (r) =

∑d|r

∑a mod N

χ(r/d, a)e2πiad/N dk,

where the inner sum is over a system of representatives for the residue classes moduloN . Interms of the Gauss sums of (2.1),

σχk (r) =

∑d|r

τ (χr/d , ζd) dk.

Finally, denote by Ak(χ) the k’th transformed Bernoulli number of (1.5) associated to thefunction χ0(n) = χ(0, n).

(2.3) Observation. Obviously, for all k ≥ 1, it follows by a change of summation order in(2.2.2) that Eχk (z+ 1) = E

φk (z) where φ(m, n) := χ(m, n− 1); in particular, it follows that

Eχk (z + N) = E

χk (z). If k ≥ 3, then it follows by a change of summation order in (2.2.1)

that Eχk (−1/z) = zkEϕk (z), where ϕ(m, n) := χ(−n,m). The argument does not work for

k = 2 or k = 1.

(2.4) Proposition. In the setup of (2.2), assume either that k ≥ 2, or that k = 1 and∑a χ(m, a) = 0 for all m. Then, with q = e2πiz/N ,

∑m≥1

∑n

χ(m, n)

(mz+ n)k= (−1)k

(k − 1)!

(2πi

N

)k ∞∑r=1

σχk−1(r)q

r . (2.4.1)

On the left, the sum over n is the symmetric summation when k = 1.

Proof. Assume that m ≥ 1. Then q(mz) = q(z)m. Hence, by (2.1.4) applied to mz,

∑n

χ(m, n)

(mz + n)k= (−1)k

(k − 1)!

(2πi

N

)k ∞∑d=1

τ(χm, ζd)dk−1qmd. (2.4.2)

On the right side, the Gauss sum is bounded as a function of m, n since χ is assumed to bebounded. Obviously, the double series

∑m,d d

k−1qmd is normally convergent when q is inthe unit disk. Hence the sum over m ≥ 1 of the functions in (2.4.2) converges normally. Inthe resulting equation, rearrange the summation on the right side according to the value ofr := md. Clearly (2.4.1) results.

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[App] 11

(2.5) Corollary. Assume either that k ≥ 2, or that k = 1 and∑

a χ(m, a) = 0 for all m.

Define χ∗(m, n) := χ(−m,−n), and let br = σχk−1(r)+ (−1)kσχ

∗k−1(r) for r ≥ 1. Then the

Fourier expansion of Eχk (z) in H is the following, with q = e2πiz/N ,

Eχk (z) = −πkAk(χ)

k!+ (−1)k

(k − 1)!

(2πi

N

)k ∑r≥1

brqr ,

where br = σχk−1(r)+ (−1)kσχ

∗k−1(r). IfAk(χ) �= 0, then the normalized seriesGχ

k obtainedby dividing Eχk by its constant term has the following expansion,

Gχk (z) = 1 + (−1)k−1k

Ak(χ)

(2i

N

)k ∑r≥1

brqr .

Proof. The left side is equal to the sum of the two series in (2.2.2). The first (constant) seriesis, by (1.8.2), equal to the constant term in the asserted Fourier expansion. The second seriesis separated into two: the sum of terms for m ≥ 1 and the sum of terms for m ≤ −1. Forthe first sum, use the equation (2.4.1). In the second sum, replace m by −m to obtain thefollowing sum: ∑

m≥1

∑n

χ(−m, n)(−mz + n)k

.

Replace n by −n in the inner sum and multiply by (−1)k . The result is the left hand side of(2.4.1) with χ replaced by χ∗. Thus (2.4.1) applies to the second sum as well, and we obtainfor the Fourier coefficient br the asserted sum.

The expression for the normalized series follows immediately from the expression forEχk (z).

(2.6) Example. Assume that χ is the constant function χ(m, n) = 1. Then N = 1, and thecorresponding function Ek is the Eisenstein series considered in (Autm.2.1) for k ≥ 3,

Ek(z) =∑′ 1

(nz +m)k.

The divisor sum is the sum,σk−1(r) =

∑d|r

dk−1.

The series vanishes when k is odd. Assume that k ≥ 2 is even. Then br = 2σk−1(r). Hencewe obtain the Fourier expansion,

Ek(z) = 2ζ(k)+ (−1)k/22(2π)k

(k − 1)!

∑r≥1

σk−1(r)qr , q = e2πiz. (2.6.1)

167

[App] 12 Automorphic functions16. februar 1995

The normalized series Gk is obtained by dividing by 2ζ(k). In the notation of Section 1,2ζ(k) = πkAk−1/(2k − 1)(k − 1)!. Hence we obtain for Gk the Fourier series,

Gk(z) = 1 + (−1)k/22k+1(2k − 1)

Ak−1

∑r≥1

σk−1(r)qr . (2.6.2)

In particular, the constant term is equal to 1 and all the Fourier coefficients are rationalnumbers. In particular,

G2(z) = 1 − 24∑r≥1

σ1(r)qr ,

G4(z) = 1 + 240∑r≥1

σ3(r)qr , (2.6.3)

G6(z) = 1 − 504∑r≥1

σ5(r)qr .

The number Ak−1 for even k is expressed by Bernoulli numbers in (1.2). It follows that thefactor in (2.6.2) is equal to −2k/Bk , where Bk is the usual Bernoulli number.

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[App] 13

3. The function of Weierstrass.

(3.1) Setup. Fix in C a lattice �, say � = Zω1 + Zω2 where the complex numbers ω1 andω2 are linearly independent over R. Denote by M(�) the field of meromorphic �-periodicfunctions, that is, a function f belongs to M(�) if and only if f is meromorphic in C andhas any ω in � as period:

f (z+ ω) = f (z).

The group� acts on C by translations, and the�-periodic functions are the functions invariantunder the action. Obviously, a fundamental domain for the action is any lattice parallelogram,

R = {z0 + t1ω1 + t2ω2 | 0 ≤ ti ≤ 1}.

The equivalence on the boundary of R identifies the opposite sides of the parallelogram.It follows that the quotient X := C/� is a torus. In particular, the quotient is a compactconnected Riemann surface of genus 1, and M(�) is the field of meromorphic functions onX.

(3.2) Lemma. Let f be a non-zero function inM(�). Then the following two formulas hold:

∑u mod �

ordu f = 0,∑

u mod �

(ordu f )u ≡ 0 (mod �). (3.2.1)

Proof. The boundary of the fundamental domain R is formed by four line segments: thesegment L1 from z0 to z0 + ω2, the segment L2 from z0 to z0 + ω1, and the two translatedsegments L′

j := Lj + ωj for j = 1, 2. We may assume that the angle from ω2 to ω1 ispositive, that is, the quotient ω1/ω2 has positive imaginary part. Then the boundary of R, asan oriented path, is the sum

∂R = L1 + L′2 − L′

1 − L2.

Clearly, we may choose z0 such that f has no zeros or poles on the boundary of R. Considerthe path integral,

1

2πi

∫∂R

f ′(z)f (z)

dz. (3.2.2)

Its value is, by the residue formula, equal to the left hand side of the first formula in (3.2.1).On the other hand, since f ′/f is �-periodic, it follows that

∫L′j(f ′/f )dz = ∫

Lj(f ′/f )dz.

Hence the path integral (3.2.2) is equal to 0. Therefore, the first formula in (3.2.1) holds.To prove the second formula, consider the path integral,

1

2πi

∫∂R

zf ′(z)f (z)

dz. (3.2.3)

The poles of the integrand are the poles and the zeros of f (z). At a point u which is eithera zero or a pole of f , the residue of zf ′(z)/f (z) is equal to ku, where k = ordu f . Hence,

169


by the residue formula, the path integral is equal to the left hand side of the second formulain (3.2.1). On the other hand, since f ′/f is �-periodic, it follows that

∫L′j(zf ′/f )dz =∫

Lj(zf ′/f )dz + ωj

∫Lj(f ′/f )dz. Hence the path integral (3.2.3) is equal to the following

sum,

− ω1

2πi

∫L1

f ′(z)f (z)

dz+ ω2

2πi

∫L2

f ′(z)f (z)

dz. (3.2.4)

By the transformation formula for integrals, the path integral∫Lj(f ′/f )dz is equal to the

integral∫fLj

(1/w)dw. Moreover, the image path fLj is a closed path in C∗. Therefore,

the integral (2πi)−1∫Lj(f ′/f )dz is an integer. It follows that the sum (3.2.4) is an integral

linear combination of ω1 and ω2. Hence the sum belongs to �. Thus the second formula of(3.2.1) holds.

(3.3) Remark. The two formulas of (3.2) are trivial if f is constant. Assume that f is anon-constant function in M(�). Then the formulas apply to f (z) − λ for any λ in C. Itfollows from the first formula that the number of times f (u), for umodulo�, takes the valueλ is equal to the number of poles; in particular, the number is independent of λ. Similarly, bythe second formula, again modulo�, the sum of the complex numbers u for which f (u) = λ

is congruent to the sum of the poles of f ; in particular, the sum is modulo� independent ofλ.

As a special case, it follows that if a function f in M(�) has no poles, then f is constant.

(3.4). The Weierstrass ℘-function is the function ℘(u) = ℘�(u) defined by the series,

℘(u) = 1

u2 +∑ω �=0

( 1

(u− ω)2− 1

ω2

), (3.4.1)

where the sum is over non-zero ω in �. The series is normally convergent in C and thefunction℘(u) is a meromorphic function with poles of order 2 at the points of�. Obviously,it is an even function: ℘(−u) = ℘(u).

It is not hard to see directly that ℘(u) is �-periodic. Alternatively, we may proceed asfollows: consider the derivative:

℘ ′(u) = −2

u3 +∑ω �=0

−2

(u− ω)3=

∑ω

−2

(u− ω)3.

Clearly, the derivative ℘ ′(u) is �-periodic. Therefore, for ω0 ∈ �, there is an equation,℘(u + ω0)− ℘(u) = C with a constant C = C(ω0). Take u := −u − ω0 in the equation.Since ℘(u) is an even function, it follows that C = 0. Hence ℘(u) is �-periodic.

To obtain the Laurent expansion of℘(u) at the origin, consider the difference℘(u)−1/u2.It follows from (3.4.1) that the difference is holomorphic at the origin with the value 0.Moreover, by applying the operator (d/du)k for k ≥ 1 to the difference, we obtain the series,

∑ω �=0

(−1)k(k + 1)!

(u− ω)k+2 .

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[App] 15

Clearly, the value of the series at the origin is equal to the number (k + 1)!Ek+2 whereEk = Ek(�) = ∑

ω �=0 ω−k as defined in (Autm.2.1). Hence ℘(u) has at the origin the

Laurent expansion,

℘(u) = 1

u2 +∑k≥1

(k + 1)Ek+2uk. (3.4.2)

From the expansions for ℘(u) and ℘ ′(u),

℘ = u−2 + 3E4u2 + 5E6u

4 + · · · , ℘ ′ = −2u−3 + 6E4u+ 20E6u3 + · · · ,

we obtain easily the equation,

(℘ ′)2 − 4℘3 + 60E4℘ = −140E6 + · · · .

The left hand side is �-periodic and its possible poles belong to �. The right hand side isholomorphic at 0. Therefore, the left hand side has no poles. It follows that the left handside is constant, equal to −140E6. Hence, the ℘-function satisfies the following differentialequation:

(℘ ′)2 = 4℘3 − g2℘ − g3, (3.4.3)

where g2 = g2(�) = 60E4(�) and g3 = g3(�) = 140E6(�).

(3.5). Modulo ω, the origin is the only pole of ℘ ′(u), and it is a pole of order 3. Therefore,by (3.2), ℘(u) has 3 zeros. Consider the three numbers u1 := ω1/2, u2 = ω2/2, andu3 := (ω1 +ω2)/2. They are inequivalent modulo�, and uj ≡ −uj (mod �). Since ℘ ′(u)is an odd function, the numbers uj are zeros of ℘′(u). Hence the uj are exactly the zeros of℘ ′(u).

The three values λj = ℘(uj ) are different. Indeed, for λ ∈ C, the function ℘(u)− λ hasa pole of order 2 at the origin. Hence, again by (3.2), the function ℘(u) − λ has two zeros.For λ = λj , the point uj is a zero for ℘(u) − λj , and of multiplicity 2, since ℘ ′(uj ) = 0.Hence uj is the only zero of ℘(u)− λj .

By (3.4.3), the values λj = ℘(uj ) are roots of the polynomial 4X3 − g2X− g3, and theyare different. The latter polynomial has, for arbitrary g2, g3 in C, three roots λj in C, and thediscriminant of the polynomial is the number,

D = 16(λ1 − λ2)2(λ1 − λ3)

2(λ2 − λ3)2 = g 3

2 − 27g 23 .

Hence, for g2 = g2(�) and g3 = g3(�), it follows that the discriminantD(�) = g 32 − 27g 2

3is non-zero.

(3.6) Proposition. Let g2 = g2(�) and g3 = g3(�). Then the discriminant D(�) =g 3

2 − 27g 23 is non-zero. Moreover, the map u �→ (℘ (u), ℘ ′(u)) induces a bijection from the

set of points u in C −� modulo � to the set of pairs (x, y) ∈ C2 satisfying the equation

y2 = 4x3 − g2x − g3. (3.6.1)

171


Proof. The first assertion was proved in (3.5). To prove the second, note first that, by (3.4.3),the pair (x, y) := (℘ (u), ℘ ′(u)) satisfies the equation (3.6.1).

Consider conversely a pair (x0, y0) satisfying the equation (3.6.1). The function℘(u)−x0has, counted with multiplicity, two zeros. If u0 is a zero, then so is −u0. If y0 = 0, thenx0 is one of the three roots λj in the polynomial 4x3

0 − g2 − g3 and u0 is, modulo �, thecorresponding unique zero uj of ℘(u) − λj . If y0 �= 0, then u0 is not one of the numbersuj . Hence the two numbers u0 and −u0 are different modulo �. Moreover, the two values℘ ′(u0) and ℘ ′(−u0) = −℘ ′(u0) are non-zero and hence different. Hence exactly one of thevalues is equal to y0 and the other value is equal to −y0.

Therefore, in both cases, there is a unique u0 modulo � such that (℘ (u0), ℘′(u0)) =

(x0, y0).

(3.7) Note. Let z be a point in the upper half plane H. Consider the lattice �z := Zz + Z.Then there is an associated ℘-function ℘�z(w). Moreover, the numbers Ek(�z), for k ≥ 3,and g2(�z) and g3(�z), are functions of z. In fact, they are the functions Ek(z) of Example(Autm.2.1), and g2(z) and g3(z) of Example (Autm.2.5).

By Example (2.5),Ek(z) = 2ζ(k)Gk(z) and 2ζ(k) = πkAk/(2k−1)(k−1)!. In particular,as A4 = 2 and A6 = 24,

g2(z) = 60E4(z) = (2π)4

22 · 3G4(z), g3(z) = 140E4(z) = (2π)6

23 · 33G6(z).

As a consequence,

D(z) = g2(z)3 − 27g3(z)

2 = (2π)12 · G4(z)3 −G6(z)

2

123 .

Hence, except for the factor (2π)12, the discriminant of (3.6), as a function D(z) of z ∈ H,is equal to the discriminant �(z) of Example (Autm.2.3), cf. (Autm.7.1). Thus the non-vanishing of D(�) implies the non-vanishing of �(z). Similarly,

g2(z)3

g2(z)3 − 27g3(z)2= G4(z)

3

G4(z)3 − 27G6(z)2. (3.7.1)

It follows that the left hand side of (3.7.1) multiplied by 123 is equal to Klein’s function j (z)defined in (Autm.2.5).

In general, for a pair (a2, a3) of complex numbers such that a 32 − 27a 2

3 �= 0, we will write

j (a2, a3) := 123a 32

a 32 − 27a 2

3

.

If (a′2, a

′3) is a second pair, then j (a′

2, a′3) = j (a2, a3) if and only if, for some non-zero

number λ,a′

2 = λ4a2, a′3 = λ6a3. (3.7.2)

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[App] 17

Indeed, the “if” part is obvious. Assume conversely the equality of the two numbers j .Clearly, if a2 = 0, then a′

2 = 0, and so (3.7.2) holds with 6 possible values of λ. Assumethat a2 �= 0. Then the first equation of (3.7.2) can be solved with 4 values of λ. For any suchλ, it follows from the equality of the two j ’s that a 2

3 = λ12a′ 23 . Therefore, if a3 �= 0 then the

two equations (3.7.2) hold for 4 values of λ, and if a3 = 0 then the equations (3.7.2) hold for2 values of λ.

(3.8) Proposition. Given two complex numbers a2 and a3 such that a 32 − 27a 2

3 �= 0. Thenthere exists a unique lattice � in C such that a2 = g2(�) and a3 = g3(�).

Proof. Note first that any lattice � is of the form,

� = λ�z, (3.8.1)

for someλ ∈ C∗ and z ∈ H. Indeed,� = Zω1+Zω2, where we may assume thatω1/ω2 ∈ H;hence it suffices to take λ := ω2 and z := ω1/ω2. Moreover, z is uniquely determined up tomultiplication by a matrix in SL2(Z).

Next, note that the functions g2(�) and g3(�) are homogeneous, respectively of degree−4 and −6:

g2(λ�) = λ−4g2(�), g3(λ�) = λ−6g3(�). (3.8.2)

Indeed, more generally, it is obvious that Ek(�) is homogeneous of degree −k.Now, by (3.7), Klein’s j -invariant is the function j (z) = j (g2(z), g3(z)). The j -invariant

is, by (Autm.7.4), an isomorphism H/ (1) ∼−→ C. In particular, there is a unique orbit (1)z in H such that j (a2, a3) = j (z). It follows, as noted in (3.7), that a2 = λ−4g2(z)

and a3 = λ−6g3(z). Therefore, by (3.8.2), if � is defined by (3.8.1), then a2 = g2(�) anda3 = g3(�).

To prove that � is unique, assume for a second lattice �′ the following two equations:

g2(�) = g2(�′), g3(�) = g3(�

′). (3.8.3)

The equations are preserved if � and �′ are multiplied by the same factor. Hence we mayassume that

� = λ�z �′ = �z′, (3.8.4)

for a non-zero λ and z, z′ ∈ H. In fact we may assume that z and z′ belong to a given systemof representatives in the standard fundamental domain F for the action of SL2(Z) on H.

From the equations (3.8.4), it follows that j (z) = j (z′). Therefore, since z and z′ areassumed to belong to a system of representatives, it follows that z = z ′. Hence,

� = λ�z, �′ = �z. (3.8.5)

Therefore, by (3.8.2), (3.8.3), and (3.8.5),

g2(z) = λ4g2(z), g3(z) = λ6g3(z). (3.8.6)

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If g2(z) and g3(z) are non-zero, it follows that λ = ±1; in particular, then �z is invariantunder multiplication by λ. If g2(z) = 0, then it follows from the second equation that λ isa sixth root of unity. Moreover, since z is a zero of g2, it is a zero of G4, and hence, by(Autm.7.3), z = ρ. Therefore, the lattice �z is invariant under multiplication by λ. Finally,if g3(z) is equal to 0, it follows similarly that λ is a fourth root of unity and that z = i;hence �z is invariant under multiplication by λ. Thus, in all cases, the lattice �z is invariantunder multiplication by λ. Therefore, by (3.8.5),� = �′, and the proof of the Proposition iscomplete.

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[App] 19

4. Notes on elliptic curves.

(4.1) Note. By definition, an elliptic curve is a pair (X, o) consisting of a (compact, connected)Riemann surface X and a distinguished point o of X. For instance, the torus X := C/�associated with a lattice � in (3.1), with the image of 0 as distinguished point, is an ellipticcurve. As a second example, let g2 and g3 be complex numbers such the g 3

2 − 27g 23 �= 0.

Consider the equation,y2 = 4x3 − g2x − g3. (4.1.1)

The equation defines an affine algebraic curve in the affine plane C2. The affine plane C2

is an open subset of the projective plane IP 2(C). In fact, the points of IP 2(C) are definedby homogeneous non-zero sets of coordinates (x, y, z) up to multiplication by a (non-zero)scalar. Among these sets of coordinates are the equivalence classes of coordinates (x, y, z)for which z �= 0. Clearly, the latter equivalence classes are represented by coordinates ofthe form (x, y, 1), and hence by points (x, y) of C2. The projective curve corresponding tothe equation (4.1.1) is the set of points in IP 2(C) whose homogeneous coordinates (x, y, z)satisfy the homogenized equation,

y2z = 4x3 − g2xz2 − g3z

3. (4.1.2)

Clearly, the homogenized equation (4.1.2) is satisfied for (x, y, z) with z �= 1 if and onlyif (4.1.1) is satisfied for (x/z, y/z). In addition, the homogeneous equation is satisfied for(x, y, 0) if and only if x = 0. Hence the projective curve consists of the points of the affinecurve and the point represented by (0, 1, 0).

The projective curve defined by (4.1.2) will be denoted X(g2, g3). It is not hard to provethat X(g2, g3) is a Riemann surface.

By Proposition (3.8), there is a lattice � such that g2 = g2(�) and g3 = g3(�). Let℘(u) = ℘�(u) be the function of Weierstrass. It follows from Proposition (3.6) that the mapu �→ (℘ (u), ℘ ′(u), 1) for u /∈ � and u �→ (0, 1, 0) for u ∈ � induces a bijection,

C/� ∼−→X(g2, g3).

Clearly, the map is a map of Riemann surfaces. Hence it is an isomorphism. In particular,therefore X(g2, g3) is a Riemann surface of genus 1, and with o := (0, 1, 0) as distinguishedpoint, it is an elliptic curve.

(4.2) Note. LetX be a Riemann surface. The divisors onX, see (Autm.8.3), form an abeliangroup, freely generated by the points of X. Two linearly equivalent divisors have the samedegree. The group of linear equivalence classes of divisors is the Picard group PicX. Denoteby PicnX the set of equivalence classes of degree n. Then each PicnX is a coset modulo thesubgroup Pic0X.

For any divisorD of X, denote by cl(D) the equivalence class of D in Pic(X). A point pof X defines a divisor 1.p of degree 1; its class cl(1.p) belongs to Pic1X.

175


Lemma I. LetX be a Riemann surface of genus 1. Then the map p �→ cl(1.p) is a bijection,

X ∼−→ Pic1X.

Proof. Since the genus is equal to 1, it follows from Riemann-Roch’s Theorem that

dimH 0(D) = degD if degD > 0. (4.2.1)

We have to prove the following assertion: for any divisor D of degree 1, there is a uniquepoint p of X such that D and 1.p are linearly equivalent, that is, there is a unique point p ofX such that, for some non-zero ϕ ∈ M(X),

divϕ +D = 1.p . (4.2.2)

Now, the left hand side divϕ +D is a divisor of degree 1. Hence it is of the form 1.p if andonly if it is positive: divϕ + D ≥ 0. By (4.2.1), the functions ϕ for which divϕ + D ≥ 0form a one-dimensional subspace of M(X). Therefore, the assertion holds.

(4.3) Note. Consider an elliptic curve (X, o). Clearly, the map D �→ D + 1.o inducesa bijection Pic0X ∼−→ Pic1X. By composing with the bijection of Lemma I, we obtain abijection from X to the commutative group Pic0X. Hence there is a unique structure as anabelian group on the points of X such that the bijection is an isomorphism of groups. Thestructure will be called the elliptic addition on X, and it will be denoted additively. Clearly,under the bijection, the distinguished point o of X corresponds to the zero element of Pic0X.Hence o is the zero element of the elliptic addition.

Unwinding the definition, the elliptic addition of two points p and q on X is the uniquepoint r = p + q of X such that we have the linear equivalence of divisors,

(1.p − 1.o)+ (1.q − 1.o) ≡ 1.r − 1.o ,

or, equivalently,1.p + 1.q ≡ 1.r + 1.o . (4.2.3)

Lemma II. The elliptic addition, as a group structure on (X, o), is characterized by thefollowing property: Let ϕ be a non-zero meromorphic function on X. Then the sum,∑p∈X(ordp ϕ)p, with respect to the elliptic addition, is equal o.

Proof. Let us first prove that the property characterizes the elliptic addition. Clearly, by theproperty applied to a constant function, the element o is the zero element. Ifp and q are pointsofX, let r be the (unique) point defined by the relation (4.2.3). The latter relation means thatfor some non-zero function ϕ we have that divϕ = 1.r + 1.o− 1.p− 1.q. Therefore, by theproperty, the equation r + o− p − q = o holds in the group X. It follows that r = p + q.

Next we prove that the property holds: By construction of the elliptic addition, the mapp �→ cl(1.p − 1.o) is a group isomorphism from X to Pic0X. Thus is suffices to prove thatthe sum,

∑p∈X(ordp ϕ)p, with respect to the elliptic addition, under this isomorphism goes

to the zero class of Pic0X. Clearly, the sum goes to the class of the divisor,∑p∈X

(ordp ϕ)(1.p − 1.o) = divϕ − (deg div ϕ).o .

The divisor divϕ is principal and its degree is equal zero. Hence the class of the divisor isequal to the zero class. Thus the elliptic addition has the property.

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[App] 21

(4.4) Note. Consider the elliptic curve X = C/� of (4.1). It has a natural group structureas a quotient of the additive group of C modulo the subgroup �. It follows from Lemma IIabove and Lemma (3.2) that the addition induced on X by the addition of C is equal to theelliptic addition.

Consider the elliptic curve X(g2, g3) of (4.1). It has a group structure defined by the wellknown addition on a smooth cubic. The latter addition is defined as follows: consider, tosimplify, two different points p = (x1, y1) and q = (x2, y2) on the affine part of the curve.Let L = 0 be the equation of the line through p and q. It intersects the curve in a third point son the curve. Let M = 0 be the equation of the line through o and s (it is simply the verticalline through s). By definition, the third point r of intersection of M = 0 and the curve is thecomposition of p and q. Now, the quotient L/M may be viewed as a meromorphic functionof the curve. It is not hard to see that

div(L/M) = 1.p + 1.q + 1.s − (1.o+ 1.s + 1.r) = 1.p + 1.q − 1.o− 1.r .

Therefore, the point r is also the elliptic sum of p and q.

(4.5) Note. Let (X, o) be an elliptic curve. We will sketch the proof that (X, o) is isomorphicto a curve of the form X(g2, g3) (the Weierstrass normal form).

First, since X is of genus 1, it follows from the Theorem of Riemann-Roch that H 0(n.o)

is of dimension n for n > 0. In particular, the following relations hold:

H 0(0.o) = H 1(1.o) ⊂ H 0(2.o) ⊂ H 0(3.o) ⊂ H 0(4.o) ⊂ H 0(5.o) ⊂ H 0(6.o).

(The first equality holds becauseH 0(0) is the 1-dimensional space of constant functions andcontained in the 1-dimensional space H 0(1.o).)

Now choose a function x in H 0(2.o) and not in H 0(1.o) and a function y in H 0(3.o)and not in H 0(2.o). The functions x and y are holomorphic except at o; at o they have,respectively, a pole of order at most 2 and at most 3. Since x /∈ H 0(1.o), it follows thatordo x = −2. Similarly, ordo y = −3. Clearly, the function x2 belongs to H 0(4.o) andsince ordo x2 = −4 it does not belong to H 0(3.o). Similarly, xy belongs to H 0(5.o) and itdoes not belong to H 0(4.o), and the two functions x3 and y2 belong to H 0(6.o) and not toH 0(5.o). It follows that the 5 functions 1, x, y, x2, xy, form a basis for H 0(5.o), and thatthey supplemented with any of the two functions y2 or x3 form a basis of H 0(6.o). Hence,there is a relation,

y2 = ax3 + bxy + cx2 + dy + ex + f,

with a �= 0. Now, in the choice of x and y, we could replace x by αx + β for α �= 0, and wecould replace y by γy+δx+ε for γ �= 0. It is not hard to see that with suitable replacements,we can obtain an equation in the Weierstrass normal form,

y2 = 4x3 − g2x − g3, (4.5.1)

and, moreover, the coefficients g2 and g3 are uniquely determined up to a choice of the formg2 �→ λ4g2 and g3 �→ λ6g3. The two functions x and y are holomorphic on the complement

177


of o. Hence they define a map from X − {o} into the subset of C2 defined by the equation(4.5.1).

The function y has a pole of order 3 at o and no more poles. Therefore, the function y hasthree zeros p1, p2, p3, not necessarily different. By the property of Lemma II, in the groupof X,

p1 + p2 + p3 = o . (4.5.2)

Consider the polynomial on the right side of Equation (4.5.1). It follows from the equationthat the value λ1 := x(p1) is a root of the polynomial. The function x − λ1 has a pole oforder 2 at o and no more poles. Hence it has two zeros, one of which is of course p1. Let p′

1be the other zero (not necessarily different form p1). Then, by the property of Lemma II,

p1 + p′1 = o . (4.5.3)

By comparing (4.5.3) and (4.5.2), it follows that p ′1 is different from p2 and p3. On the other

hand, since λ1 = x(p′1) is root of the polynomial, it follows from (4.5.2) that p ′

1 is a zeroof y. Thus p′

1 is one of the points p1, p2, p3 and since p′1 is different from p2 and p3, we

have necessarily p′1 = p1. Hence p1 is different from p2 and p3. We conclude that the 3

points p1, p2 and p3 are different and that the function x takes 3 different values on thesethree points. As the latter values are roots of the polynomial, it follows that the discriminantg 3

2 − 27g 23 of the polynomial is non-zero.

Hence the elliptic curve X(g2, g3) is defined, and we have obtained a map from X toX(g2, g3). It follows as in the proof of (3.6) that the map is bijective. Hence it is anisomorphism of Riemann surfaces and of elliptic curves.

(4.6) Note. Given an elliptic curve (X, o). By (4.5), (X, o) is isomorphic to the elliptic curveX(g2, g3). Moreover, (g2, g3) are unique up to the transformation defined in (4.5). It followsthat the following complex number,

j (X, o) := 123g 32

g 32 − 27g 2

3

.

is a well defined invariant of the elliptic curve. The invariant characterizes the elliptic curve,that is, two curves (X, o) and (X′, o′) are isomorphic if and only if j (X, o) = j (X′, o′).Indeed, assume that the two invariants are the same. Then, by (3.7), with an obvious notation,we have the equations g′

2 = λ4g2 and g′3 = λ6g3 with a non-zero scalar λ. Hence the two

curves are isomorphic.As noted in (4.1), every elliptic curve X(g2, g3) has a parameterization x = ℘(u), y =

℘ ′(u) with a Weierstrass ℘-function defined by a suitable lattice �. Therefore, it followsfrom (4.5) that every abstract elliptic curve (X, o) is isomorphic to torus C/�.

(4.7) Exercise. Prove the addition formula for the ℘-function: Let u, v and w be complexnumbers in the complement of �. Assume that u+ v + w belongs to �. Then,∣∣∣∣∣

℘(u) ℘ ′(u) 1℘(v) ℘ ′(v) 1℘(w) ℘ ′(w) 1

∣∣∣∣∣ = 0.

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[Ind] 1

I. Index.

angle, Discr.3.3angular sectors, Discr.3.3anti-transformation, Möb.1.11automorphic action, Autm.1.1automorphic factor, Autm.1.1automorphic form, Autm.4.8automorphic function, Autm.4.8automorphy equation, Autm.1.1axis, Möb.3.13balanced, Möb.1.3Bernoulli numbers, App.1.1, 1.5boundary segments, Discr.3.2boundary transformation, Discr.3.4canonical divisor, Autm.8.3canonical domain, Discr.4.1canonical generator, Discr.1.3canonical generator, Discr.1.4Cayley transformation, Möb.1.5chain, Autm.8.1congruence subgroup, Mdlar.3.1conjugate, Autm.3.10conjugate, Autm.3.16conjugation, Möb.2.6cross ratio, Möb.1.6cusp form, Autm.4.8cusp, Discr.1.4cusp, Discr.3.2cuspidal divisor, Autm.8.6degree, Autm.8.3determination, Autm.1.8discriminant, Autm.2.3disk, Möb.1.8divisor, Autm.8.3dual factor, Autm.8.10η-factor, Autm.2.3η-function, Autm.2.3Eisenstein series, App.2.2Eisenstein series, Autm.2.1Eisenstein series, Poinc.1.4

elliptic addition, App.4.3elliptic curve, App.4.1elliptic matrix, Möb.2.1elliptic orbit, Discr.1.4elliptic point, Discr.2.8elliptic point, Möb.2.7elliptic side, Discr.5.1Euler characteristic, Discr.5.1Euler characteristic, Discr.5.2Euler numbers, App.1.1Euler zigzag numbers, App.1.2exponentially bounded, Autm.4.1exponentially bounded, Autm.4.4factor, Autm.1.9finite disk, Autm.1.7finite domain, Discr.3.2first kind, Discr.3.15Fourier coefficients, Autm.5.2Fourier expansion, Autm.5.2Fuchsian group, Discr.3.15fundamental domain, Discr.3.1fundamental neighborhood, Discr.2.8 -order, Autm.4.5Gauss sum, App.2.1genus, Autm.8.3genus, Discr.5.1geodesic circle, Möb.3.6half turn, Möb.3.15homogeneous factors, Autm.1.11homogeneous, Möb.1.4homogenized group, Mdlar.3.1horo cycle, Discr.2.8hyperbolic matrix, Möb.2.1inhomogeneous, Möb.1.4integral form, Autm.4.8invariant, Autm.1.9irregular cusp, Discr.1.4j -automorphic function, Autm.4.8j -regular cusp, Autm.6.10

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[Ind] 2 Automorphic functions16. februar 1995

j -invariant, Autm.2.5Klein’s j -invariant, Autm.2.5level-N subgroup, Mdlar.3.1limit point, Möb.3.2limit rotation, Möb.3.13line in D, Möb.3.2linearly equivalent, Autm.8.3local parameter, Autm.5.2loxodromic, Möb.2.1Möbius transformation, Möb.1.1modular group, Mdlar.1.1nearest point, Discr.4.4non-euclidean circle, Möb.3.6non-euclidean distance, Möb.3.3normal domain, Discr.3.10order, Autm.4.1order, Discr.1.4ordinary point, Möb.2.7ordinary point, Discr.2.8parabolic matrix, Möb.2.1parabolic orbit, Discr.1.4parabolic point, Möb.2.7parameter, Autm.3.3periodic, App.3.1Picard group, App.4.2Poincaré series, Poinc.1.2principal determination, Autm.1.8principal divisor, Autm.8.3projective curve, App.4.1projective line, Mdlar.1.1proper map, Möb.4.1properly discontinuous, Discr.2.3Ramanujan’s τ -function, Autm.7.1

reflection, Möb.3.13regular cusp, Discr.1.4representatives, Möb.1.1rotation, Möb.3.13semi-direct product, Autm.1.3side, Discr.3.4side, Discr.4.3sign, Autm.3.3special number, App.1.6surface, Discr.5.1θ -factor, Autm.2.6θ -function, Autm.2.6θ -group, Mdlar.3.3τ -function, Autm.7.1Theorem A, Autm.6.2Theorem B, Autm.6.3Theorem C, Autm.6.7Theorem D, Autm.6.11Theorem E, Autm.6.15Theorem F, Autm.6.16Theorem G, Poinc.1.7Theorem H, Autm.8.9transformed Bernoulli numbers, App.1.5translation, Möb.3.13up/down numbers, App.1.2value, Autm.1.8vertex, Discr.3.2vertex, Discr.4.3Weierstrass ℘-function, App.3.4weight-k, Autm.3.16width, Discr.3.16width, Discr.4.4zigzag numbers App.1.2

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[Ind] 3

Special symbols:

Ak , special numbers, App.1.2Ak(χ) , special numbers, App.1.6Bk , Bernoulli number, App.1.1Bk(χ) , Bernoullli numbers, App.1.16D/ , closed orbit space, Discr.2.8�(z) , discriminant, Autm.2.3∂ (D) , Discr.1.1degD , degree of divisor, Autm.8.3δP , half turn, Möb.3.15divϕ , principal divisor, Autm.8.3df(u, v, w, z) , cross ratio, Möb.1.6dist , non-euclidean distance, Möb.3.3E , unit disk, Möb.1.8η(z) , Dedekind eta-function, Autm.2.3Ek , Euler number, App.1.1Ek(z) , Eisenstein series, Autm.2.1Eχk (z) , Eisenstein series, App.2.2

eu , order of elliptic point, Discr.1.4g , genus, Discr.5.1G( , j) , integral forms, Autm.4.8 (N) , congruence subgroup, Mdlar.3.1 (1) , the modular group, Mdlar.3.1Gk(z) , Eisenstein series, App.2.2Gχk (z) , Eisenstein series, App.2.5

γu , canonical generator, Discr.1.4 θ , theta group, Mdlar.3.3 0(N) , congruence subgroup, Mdlar.3.1H , upper half plane, Möb.1.8

h(α) , invariant of matrix, Möb.2.2IC(f ) , path integral, Autm.8.1J (γ, z) , denominator, Möb.1.1j (z) , Klein’s invariant, Autm.2.5jη(γ, z) , the η-factor, Autm.2.3jθ (γ, z) , the θ -factor, Autm.2.6κu , parameter, Autm.3.3μ( ) , invariant of , Discr.5.6μn , group of nth roots of unity, Mdlar.1.1M( , j) , meromorphic forms, Autm.4.8νe( ) , number of orbits, Discr.5.6ord u , the -order, Autm.4.5ωu , sign, Autm.3.3℘(u) , Weierstrass function, App.3.4PicX , Picard group, App.4.2q = e2πiz/h , local parameter, Autm.5.2ρl , reflection, Möb.3.15s , speciel matrix, Möb.1.1σχk (r) , divisor sum, App.2.2

S( , j) , cusp forms, Autm.4.8SL(D) , stabilizer of disk, Möb.1.8t , special matrix, Möb.1.1θ(z) , theta function, Autm.2.6τ(χ, ζ ) , Gauss sum, App.2.1th , translation matrix, Discr.1.2τQP , translation, Möb.3.15u , special matrix, Möb.1.1χ(X) , Euler characteristic, Discr.5.1χ4 , Dirichlet character, Poinc.3.1

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[Ind] 4 Automorphic functions16. februar 1995

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Image of front page under the Cayley tranformation.

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anders thorup notes on automorphic functionsweb.math.ku.dk/noter/filer/mdlar.pdf · automorphic...

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